{"id":624101,"date":"2023-09-01T06:45:12","date_gmt":"2023-09-01T06:45:12","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624101"},"modified":"2023-09-04T10:46:08","modified_gmt":"2023-09-04T10:46:08","slug":"kc-sinha-exercise-4-4-mathematics-solution-class-10-chapter-4-trigonometric-ratios-and-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-4-4-mathematics-solution-class-10-chapter-4-trigonometric-ratios-and-identities\/","title":{"rendered":"KC Sinha: Exercise 4.4 – Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities"},"content":{"rendered":"\n\n\n\n\n

Question 1<\/h4>\n\n\n\n

Fill in the blanks<\/strong><\/p>\n\n\n\n

(i) sin2<\/sup> \u03b8 cosec2<\/sup> \u03b8 = \u2026\u2026..<\/strong>
(ii) 1 + tan2<\/sup> \u03b8 = \u2026\u2026<\/strong>
(iii) Reciprocal sin \u03b8. cot \u03b8 = \u2026\u2026<\/strong>
(iv) 1\u2013…….= cos2<\/sup>\u03b8<\/strong>
(v) <\/strong>$\\tan A=\\frac{\\cdots \\cdots \\cdot}{\\cos A}$<\/strong>
(vi) <\/strong>$\\ldots \\ldots=\\frac{\\cos A}{\\sin A}$<\/strong>
(vii) cos \u03b8 is reciprocal of ………<\/strong>
(viii) Reciprocal of sin \u03b8 is………<\/strong>
(ix) Value of sin \u03b8 in terms of cos \u03b8 is<\/strong>
(x) Value of cos \u03b8 in terms of sin \u03b8 is<\/strong>
Sol :
(i) Given: sin2<\/sup> \u03b8 cosec2<\/sup> \u03b8<\/p>\n\n\n\n

$\\Rightarrow \\sin ^{2} \\theta \\times \\frac{1}{\\sin ^{2} \\theta}$ $\\left[\\because \\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
= 1<\/p>\n\n\n\n

(ii) Given: 1 + tan2<\/sup> \u03b8<\/p>\n\n\n\n

$=1+\\left(\\frac{\\sin \\theta}{\\cos \\theta}\\right)^{2}$ $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
$=1+\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}$
$=\\frac{\\cos ^{2} \\theta+\\sin ^{2} \\theta}{\\cos ^{2} \\theta}$[\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1}{\\cos ^{2} \\theta}$
= sec2<\/sup> \u03b8 $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$<\/p>\n\n\n\n

(iii) Given : sin \u03b8 cot \u03b8
Firstly, we simplify the given trigonometry<\/p>\n\n\n\n

$=\\sin \\theta \\times \\frac{\\cos \\theta}{\\sin \\theta}$ $\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right]$
= cos \u03b8
Now, the reciprocal of cos \u03b8 is
$=\\frac{1}{\\cos \\theta}$
=sec \u03b8 $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$<\/p>\n\n\n\n

(iv) Given: 1 \u2013 x = cos2<\/sup>\u03b8
Subtracting 1 to both the sides, we get
1 \u2013 x \u20131 = cos2<\/sup>\u03b8 \u2013 1
\u21d2 \u2013x = \u2013 sin2<\/sup>\u03b8
\u21d2 x =sin2<\/sup> \u03b8<\/p>\n\n\n\n

(v) $\\tan A=\\frac{\\sin A}{\\cos A}$<\/p>\n\n\n\n

(vi) $\\cot A=\\frac{\\cos A}{\\sin A}$<\/p>\n\n\n\n

(vii) $\\cos \\theta=\\frac{1}{\\sec \\theta}$<\/p>\n\n\n\n

(viii) $\\sin \\theta=\\frac{1}{\\cos \\theta}$<\/p>\n\n\n\n

(ix) We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 sin2<\/sup> \u03b8 = 1 \u2013 cos2<\/sup> \u03b8
\u21d2 sin \u03b8 = \u221a(1 \u2013 cos2<\/sup> \u03b8)
(x) We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 cos 2<\/sup> \u03b8 = 1 \u2013 sin2<\/sup> \u03b8
\u21d2 cos \u03b8 = \u221a(1 \u2013 sin2<\/sup> \u03b8)<\/p>\n\n\n\n

\n\n\n\n

Question 2<\/h4>\n\n\n\n

If sin \u03b8= p and cos \u03b8 = q, what is the relation between p and q ?<\/strong><\/p>\n\n\n\n

Sol :
We know that,
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1 \u2026(i)
Given : sin \u03b8 = p and cos \u03b8 = q
Putting the values of sin \u03b8 and cos \u03b8 in eq. (i), we get
(q)2<\/sup> + (p)2<\/sup> =1
\u21d2 p2<\/sup> + q2<\/sup> =1<\/p>\n\n\n\n

\n\n\n\n

Question 3<\/h4>\n\n\n\n

If cos A = x, express sin A in terms of x<\/strong><\/p>\n\n\n\n

Sol :
We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 sin2<\/sup> \u03b8 = 1 \u2013 cos2<\/sup> \u03b8
\u21d2 sin \u03b8 = \u221a(1 \u2013 cos2<\/sup> \u03b8)
And Given that cos \u03b8 = x
\u21d2 sin \u03b8 = \u221a(1\u2013 x2<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Question 4<\/h4>\n\n\n\n

If x cos \u03b8 = 1 and y sin \u03b8 = 1 find the value of tan \u03b8.<\/strong><\/p>\n\n\n\n

Sol :
Given x cos\u03b8 = 1 and y sin\u03b8 = 1<\/p>\n\n\n\n

$\\Rightarrow \\cos \\theta=\\frac{1}{x}$ and $\\sin \\theta=\\frac{1}{\\mathrm{y}}$<\/p>\n\n\n\n

Now, we know that<\/p>\n\n\n\n

$\\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}$<\/p>\n\n\n\n

Putting the value of sin \u03b8 and cos \u03b8, we get
$\\tan \\theta=\\frac{\\frac{1}{y}}{\\frac{1}{x}}$
$\\Rightarrow \\tan \\theta=\\frac{x}{y}$<\/p>\n\n\n\n

\n\n\n\n

Question 5<\/h4>\n\n\n\n

If cos40o<\/sup> = p, then write the value of sin 40o<\/sup> in terms of p.<\/strong><\/p>\n\n\n\n

Sol :
We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 cos2<\/sup> 40\u00b0 + sin2<\/sup> 40\u00b0 = 1
\u21d2 sin2<\/sup> 40\u00b0 = 1 \u2013 cos2<\/sup> 40\u00b0
\u21d2 sin 40\u00b0 = \u221a(1 \u2013 cos2<\/sup> 40\u00b0)
And Given that cos 40\u00b0 = p
\u21d2 sin 40\u00b0 = \u221a(1\u2013 p2<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Question 6<\/h4>\n\n\n\n

If sin 77\u00b0 = x, then write the value of cos 77o<\/sup> in terms of x.<\/strong><\/p>\n\n\n\n

Sol :
We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 cos2<\/sup> 77\u00b0 + sin2<\/sup> 77\u00b0 = 1
\u21d2 cos 2<\/sup> 77\u00b0 = 1 \u2013 sin2<\/sup> 77\u00b0
\u21d2 cos 77\u00b0 = \u221a(1 \u2013 sin2<\/sup> 77\u00b0)
And Given that sin 77\u00b0 = x
\u21d2 cos 77\u00b0 = \u221a(1 \u2013 x2<\/sup> )<\/p>\n\n\n\n

\n\n\n\n

Question 7<\/h4>\n\n\n\n

If cos55\u00b0 = x2<\/sup>, then write the value of sin 55o<\/sup> in terms of x.<\/strong><\/p>\n\n\n\n

Sol :
We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 cos2<\/sup> 55\u00b0 + sin2<\/sup> 55\u00b0 = 1
\u21d2 sin2<\/sup> 55\u00b0 = 1 \u2013 cos2<\/sup> 55\u00b0
\u21d2 sin 55\u00b0 = \u221a(1 \u2013 cos2<\/sup> 55\u00b0)
And Given that cos 55\u00b0 = x2<\/sup>
\u21d2 sin 55\u00b0 = \u221a{1\u2013 (x2<\/sup>)2<\/sup>}
\u21d2 sin 55\u00b0 = \u221a(1 \u2013 x4<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Question 8<\/h4>\n\n\n\n

If, sin 50\u00b0 = \u03b1 then write the value of cos 50\u00b0 in terms of \u03b1.<\/strong><\/p>\n\n\n\n

Sol :
We know that
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 cos2<\/sup> 50\u00b0 + sin2<\/sup> 50\u00b0 = 1
\u21d2 cos 2<\/sup> 50\u00b0 = 1 \u2013 sin2<\/sup> 50\u00b0
\u21d2 cos 50\u00b0 = \u221a(1 \u2013 sin2<\/sup> 50\u00b0)
And Given that sin 50\u00b0 = a
\u21d2 cos 50\u00b0 = \u221a(1 \u2013 a2<\/sup> )<\/p>\n\n\n\n

\n\n\n\n

Question 9<\/h4>\n\n\n\n

If x cos A = 1 and tan A = y, then what is the value of x2<\/sup> \u2013 y2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :
Given x cos A = 1 and tan A =y
$\\Rightarrow \\mathrm{x}=\\frac{1}{\\cos \\mathrm{A}}$ and $\\frac{\\sin A}{\\cos A}=y$ $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$ 
To find: x2<\/sup> \u2013 y2<\/sup><\/p>\n\n\n\n

Putting the values of x and y , we get
$\\left(\\frac{1}{\\cos A}\\right)^{2}-\\left(\\frac{\\sin A}{\\cos A}\\right)^{2}$
$=\\frac{1}{\\cos ^{2} A}-\\frac{\\sin ^{2} A}{\\cos ^{2} A}$
$=\\frac{1-\\sin ^{2} A}{\\cos ^{2} A}$
$=\\frac{\\cos ^{2} A}{\\cos ^{2} A}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= 1<\/p>\n\n\n\n

\n\n\n\n

Question 10<\/h4>\n\n\n\n

Prove the followings identities:<\/strong><\/p>\n\n\n\n

(1 \u2013 sin \u03b8)(1 + sin \u03b8) = cso2<\/sup> \u03b8<\/strong>
Sol :
Taking LHS = (1 \u2013 sin\u03b8)(1+ sin\u03b8)
Using identity , (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>) , we get
= (1)2<\/sup> \u2013 (sin\u03b8)2<\/sup>
= 1 \u2013 sin2<\/sup> \u03b8
= cos2<\/sup> \u03b8 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 11<\/h4>\n\n\n\n

Prove the followings identities:<\/strong><\/p>\n\n\n\n

(1 + cos \u03b8)(1 \u2013 cos \u03b8) = sin2<\/sup>\u03b8<\/strong>
Sol :
Taking LHS =(1 \u2013 cos\u03b8)(1+ cos\u03b8)
Using identity , (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>) , we get
= (1)2<\/sup> \u2013 (cos\u03b8)2<\/sup>
= 1 \u2013 cos2<\/sup> \u03b8
= sin2<\/sup> \u03b8 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 12<\/h4>\n\n\n\n

Prove the followings identities:<\/strong><\/p>\n\n\n\n

$\\frac{(1-\\cos \\theta)(1+\\cos \\theta)}{(1-\\sin \\theta)(1+\\sin \\theta)}=\\tan ^{2} \\theta$<\/p>\n\n\n\n

Sol :
Taking LHS $\\frac{(1-\\cos \\theta)(1+\\cos \\theta)}{(1-\\sin \\theta)(1+\\sin \\theta)}$
$=\\frac{(1)^{2}-(\\cos \\theta)^{2}}{(1)^{2}-(\\sin \\theta)^{2}}$ [Using identity , (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
$=\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= tan2<\/sup> \u03b8  $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 13<\/h4>\n\n\n\n

Prove the followings identities:<\/strong><\/p>\n\n\n\n

$\\frac{1}{\\sec \\theta+\\tan \\theta}=\\sec \\theta-\\tan \\theta$<\/p>\n\n\n\n

Sol :
Taking LHS $=\\frac{1}{\\sec \\theta+\\tan \\theta}$
Multiplying and divide by the conjugate of sec \u03b8 + tan \u03b8
$=\\frac{1}{\\sec \\theta+\\tan \\theta} \\times \\frac{\\sec \\theta-\\tan \\theta}{\\sec \\theta-\\tan \\theta}$
$=\\frac{\\sec \\theta-\\tan \\theta}{\\sec ^{2} \\theta-\\tan ^{2} \\theta}$ [Using identity , (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
= sec \u03b8 \u2013 tan \u03b8 [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8 ]
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 A <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

sin\u03b8. cot\u03b8 = cos \u03b8<\/strong>
Sol :
Taking LHS = sin \u03b8 cot \u03b8
$=\\sin \\theta \\times \\frac{\\cos \\theta}{\\sin \\theta}\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right]$
= cos \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 B <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

sin2<\/sup> \u03b8(1+ cot2<\/sup> \u03b8) = 1<\/strong>
Sol :<\/p>\n\n\n\n

Taking LHS = sin2<\/sup> \u03b8(1+ cot2<\/sup> \u03b8)
= sin2<\/sup> \u03b8 (cosec2<\/sup> \u03b8) [\u2235 cot2<\/sup> \u03b8 +1= cosec2<\/sup> \u03b8 ]<\/p>\n\n\n\n

$=\\sin ^{2} \\theta \\times \\frac{1}{\\sin ^{2} \\theta}$ $\\left[\\because \\sin \\theta=\\frac{1}{\\operatorname{cscec} \\theta}\\right]$
= 1
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 C <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

cos2<\/sup> A (tan2<\/sup> A+1) = 1<\/strong>
Sol :
Taking LHS = cos2<\/sup> A (tan2<\/sup> A+1)
= cos2<\/sup> \u03b8 (sec2<\/sup> \u03b8) [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8 ]<\/p>\n\n\n\n

$=\\cos ^{2} \\theta \\times \\frac{1}{\\cos ^{2} \\theta}$ $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$
= 1
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 D <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

tan4<\/sup>\u03b8 + tan2<\/sup>\u03b8 = sec4<\/sup>\u03b8 \u2013 sec2<\/sup>\u03b8<\/strong>
Sol :
Taking LHS = tan4<\/sup> \u03b8 + tan2<\/sup> \u03b8
= (tan2<\/sup> \u03b8)2<\/sup> + tan2<\/sup> \u03b8
= ( sec2<\/sup> \u03b8 \u2013 1)2<\/sup> + (sec2<\/sup> \u03b8 \u2013 1) [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8 ]
= sec4<\/sup> \u03b8 + 1 \u2013 2 sec2<\/sup> \u03b8 + sec2<\/sup> \u03b8 \u2013 1 [\u2235 (a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)]
= sec4<\/sup> \u03b8 \u2013 sec2<\/sup> \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 E <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{\\left(1+\\tan ^{2} \\theta\\right) \\sin ^{2} \\theta}{\\tan \\theta}=\\tan \\theta$
Sol :
Taking LHS $=\\frac{\\left(1+\\tan ^{2} \\theta\\right) \\sin ^{2} \\theta}{\\tan \\theta}$
$=\\frac{\\left(\\sec ^{2} \\theta\\right) \\sin ^{2} \\theta}{\\frac{\\sin \\theta}{\\cos \\theta}}$ [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]
$=\\frac{1 \\times \\sin \\theta \\times \\cos \\theta}{\\cos ^{2} \\theta}$ $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$
$=\\frac{\\sin \\theta}{\\cos \\theta}$
= tan \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 F <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}+1=\\frac{\\tan ^{2} \\theta}{\\sin ^{2} \\theta}$
Sol :
Taking LHS$=\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}+1$
$=\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1}{\\cos ^{2} \\theta}$
= sec2<\/sup>\u03b8 $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$<\/p>\n\n\n\n

Now, RHS $=\\frac{\\tan ^{2} \\theta}{\\sin ^{2} \\theta}$<\/p>\n\n\n\n

$=\\frac{\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}}{\\sin ^{2} \\theta}$ $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$<\/p>\n\n\n\n

$=\\frac{\\sin ^{2} \\theta}{\\sin ^{2} \\theta \\times \\cos ^{2} \\theta}$<\/p>\n\n\n\n

$=\\frac{1}{\\cos ^{2} \\theta}$
= sec2<\/sup>\u03b8  $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 G <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{3-4 \\sin ^{2} \\theta}{\\cos ^{2} \\theta}=3-\\tan ^{2} \\theta$
Sol :
Taking LHS $=\\frac{3-4 \\sin ^{2} \\theta}{\\cos ^{2} \\theta}$
$=\\frac{3}{\\cos ^{2} \\theta}-\\frac{4 \\sin ^{2} \\theta}{\\cos ^{2} \\theta}$
= 3 sec2<\/sup> \u03b8 \u2013 4 tan2<\/sup> \u03b8
We know that,
1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8
= 3(1+ tan2<\/sup> \u03b8) \u2013 4 tan2<\/sup> \u03b8
= 3 + 3 tan2<\/sup> \u03b8 \u2013 4 tan2<\/sup> \u03b8
= 3 \u2013 tan2<\/sup> \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 H <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

(1+ tan2<\/sup> \u03b8) cos \u03b8. sin \u03b8 = tan \u03b8<\/strong>
Sol :
Taking LHS = (1+ tan2<\/sup> \u03b8) cos \u03b8 sin \u03b8
= (sec2<\/sup> \u03b8) cos \u03b8 sin \u03b8 [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]<\/p>\n\n\n\n

$=\\frac{1}{\\cos ^{2} \\theta} \\times \\cos \\theta \\times \\sin \\theta$ $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$
$=\\frac{\\sin \\theta}{\\cos \\theta}$
= tan \u03b8 $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 I <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

sin2<\/sup>\u03b8 \u2013 cos2<\/sup> \u03d5 = sin2<\/sup>\u03d5 \u2013 cos2<\/sup>\u03b8<\/strong>
Sol :
Taking LHS = sin2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03c6
=( 1 \u2013 cos2<\/sup> \u03b8) \u2013 (1 \u2013 sin2<\/sup> \u03c6) [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1] & [\u2235 cos2<\/sup> \u03c6 + sin2<\/sup> \u03c6 = 1]
= 1 \u2013 cos2<\/sup> \u03b8 \u2013 1 + sin2<\/sup> \u03c6
= sin2<\/sup> \u03c6 \u2013 cos2<\/sup> \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 J <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{1-\\tan ^{2} \\theta}{\\cot ^{2} \\theta-1}=\\tan ^{2} \\theta$<\/p>\n\n\n\n

Sol :
Taking LHS $=\\frac{1-\\tan ^{2} \\theta}{\\cot ^{2} \\theta-1}$
$=\\frac{1-\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}}{\\frac{\\cos ^{2} \\theta}{\\sin ^{2} \\theta}-1}$
$=\\frac{\\frac{\\cos ^{2} \\theta-\\sin ^{2} \\theta}{\\cos ^{2} \\theta}}{\\frac{\\cos ^{2} \\theta-\\sin ^{2} \\theta}{\\sin ^{2} \\theta}}$
$=\\frac{\\cos ^{2} \\theta-\\sin ^{2} \\theta}{\\cos ^{2} \\theta} \\times \\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta-\\sin ^{2} \\theta}$
= tan2<\/sup> \u03b8 $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 A <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

(1 \u2013 cos\u03b8)(1+ cos\u03b8)(1+ cot2<\/sup> \u03b8) = 1<\/strong>
Sol :
Taking LHS = (1 \u2013 cos\u03b8)(1+ cos\u03b8)(1+ cot2<\/sup> \u03b8)
Using identity , (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>) in first two terms , we get
= (1)2<\/sup> \u2013 (cos\u03b8)2<\/sup> (cosec2<\/sup> \u03b8) [\u2235 cot2<\/sup> \u03b8 +1= cosec2<\/sup> \u03b8]
= (1 \u2013 cos2<\/sup> \u03b8) (cosec2<\/sup> \u03b8)
= (sin2<\/sup> \u03b8) (cosec2<\/sup> \u03b8) [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]<\/p>\n\n\n\n

$=\\sin ^{2} \\theta \\times \\frac{1}{\\sin ^{2} \\theta}$ $\\left[\\because \\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
=1
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 B <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{(1+\\sin \\theta)^{2}+(1-\\sin \\theta)^{2}}{2 \\cos ^{2} \\theta}=\\frac{1+\\sin ^{2} \\theta}{1-\\sin ^{2} \\theta}$
Sol :
Taking LHS $=\\frac{(1+\\sin \\theta)^{2}(1-\\sin \\theta)^{2}}{2 \\cos ^{2} \\theta}$
$=\\frac{1+\\sin ^{2} \\theta+2 \\sin \\theta+1+\\sin ^{2} \\theta-2 \\sin \\theta}{2 \\cos ^{2} \\theta}$
[\u2235 (a + b)2<\/sup> = (a2<\/sup> + b2<\/sup> + 2ab) and (a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)]
$=\\frac{2+2 \\sin ^{2} \\theta}{2 \\cos ^{2} \\theta}$
$=\\frac{2\\left(1+\\sin ^{2} \\theta\\right)}{2\\left(1-\\sin ^{2} \\theta\\right)}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1+\\sin ^{2} \\theta}{1-\\sin ^{2} \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 C <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{\\cos ^{2} \\theta(1-\\cos \\theta)}{\\sin ^{2} \\theta(1-\\sin \\theta)}=\\frac{1+\\sin \\theta}{1+\\cos \\theta}$
Sol :
Taking LHS $=\\frac{\\cos ^{2} \\theta(1-\\cos \\theta)}{\\sin ^{2} \\theta(1-\\sin \\theta)}$<\/p>\n\n\n\n

Multiplying and divide by the conjugate of (1 \u2013 sin\u03b8), we get
$=\\frac{\\cos ^{2} \\theta(1-\\cos \\theta)}{\\sin ^{2} \\theta(1-\\sin \\theta)} \\times \\frac{(1+\\sin \\theta)}{(1+\\sin \\theta)}$
$=\\frac{\\cos ^{2} \\theta(1-\\cos \\theta)(1+\\sin \\theta)}{\\sin ^{2} \\theta\\left[(1)^{2}-(\\sin \\theta)^{2}\\right]}$
$=\\frac{\\cos ^{2} \\theta(1-\\cos \\theta)(1+\\sin \\theta)}{\\sin ^{2} \\theta \\times \\cos ^{2} \\theta}$
$=\\frac{(1-\\cos \\theta)(1+\\sin \\theta)}{\\sin ^{2} \\theta}$<\/p>\n\n\n\n

Now, multiply and divide by conjugate of 1 \u2013 cos \u03b8, we get
$=\\frac{(1-\\cos \\theta)(1+\\sin \\theta)}{\\sin ^{2} \\theta} \\times \\frac{(1+\\cos \\theta)}{(1+\\cos \\theta)}$
$=\\frac{\\left(1^{2}-\\cos ^{2} \\theta\\right)(1+\\sin \\theta)}{\\sin ^{2} \\theta(1+\\cos \\theta)}$
$=\\frac{\\sin ^{2} \\theta(1+\\sin \\theta)}{\\sin ^{2} \\theta(1+\\cos \\theta)}$
$=\\frac{1+\\sin \\theta}{1+\\cos \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 D <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

(sin \u03b8 \u2013 cos \u03b8)2<\/sup> = 1 \u2013 2 sin\u03b8 . cos \u03b8<\/strong>
Sol :
Taking LHS = (sin \u03b8 \u2013 cos \u03b8)2<\/sup>
Using the identity,(a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)
= sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 \u2013 2sin \u03b8 cos \u03b8
= 1 \u2013 2sin \u03b8 cos \u03b8 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 E <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

(sin \u03b8 + cos \u03b8)2<\/sup> + (sin \u03b8 \u2013 cos \u03b8)2<\/sup> = 2<\/strong>
Sol :
Taking LHS = (sin \u03b8 + cos \u03b8)2<\/sup> + (sin \u03b8 \u2013 cos \u03b8)2<\/sup>
Using the identity,(a + b)2<\/sup> = (a2<\/sup> + b2<\/sup> + 2ab) and (a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)
= sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 + 2sin \u03b8 cos \u03b8 + sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 \u2013 2sin \u03b8 cos \u03b8
= 1 +1 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= 2
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 F <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

(asin \u03b8 + bcos \u03b8)2<\/sup> + (acos \u03b8 \u2013 bsin \u03b8)2<\/sup> = a2<\/sup> + b2<\/sup><\/strong>
Sol :
Taking LHS = (asin \u03b8 + bcos \u03b8)2<\/sup> + (acos \u03b8 \u2013 bsin \u03b8 )2<\/sup>
Using the identity,(a + b)2<\/sup> = (a2<\/sup> + b2<\/sup> + 2ab) and (a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)
= a2<\/sup> sin2<\/sup> \u03b8 + b2<\/sup> cos2<\/sup> \u03b8 + 2 ab sin \u03b8 cos \u03b8 + a2<\/sup> cos2<\/sup> \u03b8 + b2<\/sup> sin2<\/sup> \u03b8 \u2013 2 ab sin \u03b8 cos \u03b8
= a2<\/sup> sin2<\/sup> \u03b8+ a2<\/sup> cos2<\/sup> \u03b8 + b2<\/sup> sin2<\/sup> \u03b8 + b2<\/sup> cos2<\/sup> \u03b8
= a2<\/sup> (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) + b2<\/sup> (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8)
= a2<\/sup> + b2<\/sup> [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 G <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

cos4<\/sup> A + sin4<\/sup> A + 2 sin2<\/sup> A. cos2<\/sup> A = 1<\/strong>
Sol :
Taking LHS = cos4<\/sup> A + sin4<\/sup> A + 2 sin2<\/sup> A cos2<\/sup> A
Using the identity,(a + b)2<\/sup> = (a2<\/sup> + b2<\/sup> + 2ab)
Here, a = cos2<\/sup> A and b = sin2<\/sup> A
= ( cos2<\/sup> A + sin2<\/sup> A) [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= 1<\/p>\n\n\n\n

\n\n\n\n

Question 15 H <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

sin4<\/sup> A \u2013 cos4<\/sup> A = 2 sin2<\/sup> A \u2013 1 = 1 \u2013 2 cos2<\/sup> A = sin2<\/sup> A \u2013 cos2<\/sup> A<\/strong>
Sol :
Given:
$\\begin{matrix}\\sin ^{4} A-\\cos ^{4} A&=2 \\sin ^{2} A-1&=1-2 \\cos ^{2} A&=\\sin ^{2} A-\\cos ^{2} A\\\\ \\text{I}&\\text{II}&\\text{III}&\\text{IV}\\end{matrix}$
Taking I term
= sin4<\/sup> A \u2013 cos4<\/sup> A \u2192 I term
= (sin2<\/sup> A)2<\/sup> \u2013 (cos2<\/sup> A)2<\/sup>
= (sin2<\/sup> A \u2013 cos2<\/sup> A)(sin2<\/sup> A+ cos2<\/sup> A )
[\u2235 (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)]
= (sin2<\/sup> A \u2013 cos2<\/sup> A)(1) [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= (sin2<\/sup> A \u2013 cos2<\/sup> A) \u2026(i) \u2192 IV term
From Eq. (i)
= {sin2<\/sup> A \u2013 (1 \u2013 sin2<\/sup> A)} [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= sin2<\/sup> A \u2013 1 + sin2<\/sup> A
= 2 sin2<\/sup> A \u2013 1 \u2192 II term
Again, From Eq. (i)
= {(1 \u2013 cos2<\/sup> A) \u2013 cos2<\/sup> A } [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=1 \u2013 2 cos2<\/sup> A \u2192 III term
Hence, I = II = III = IV
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 I <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

cos4<\/sup> \u03b8 \u2013 sin4<\/sup> \u03b8 = cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8 = 2 cos2<\/sup> \u03b8 \u2013 1<\/strong>
Sol :
Given:<\/p>\n\n\n\n

$\\begin{matrix}\\cos ^{4} \\theta-\\sin ^{4} \\theta&=\\cos ^{2} \\theta-\\sin ^{2} \\theta&=2 \\cos ^{2} \\theta-1\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n

Taking I term
= cos4<\/sup> \u03b8 \u2013 sin4<\/sup> \u03b8 \u2192 I term
= (cos2<\/sup> \u03b8)2<\/sup> \u2013 (sin2<\/sup> \u03b8)2<\/sup>
= (cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8)(cos2<\/sup> \u03b8+ sin2<\/sup> \u03b8 )
[\u2235 (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)]
= (cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8) (1) [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= (cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8) \u2026(i) \u2192 II term
From Eq. (i)
= {cos2<\/sup> \u03b8 \u2013 (1 \u2013 cos2<\/sup> \u03b8)} [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= 2 cos2<\/sup> \u03b8 \u2013 1 \u2192 III term
Hence, I = II = III
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 J <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

2 cos2<\/sup> \u03b8 \u2013 cos4<\/sup> \u03b8 + sin4<\/sup> \u03b8 = 1<\/strong>
Sol :
Taking LHS = 2 cos2<\/sup> \u03b8 \u2013 cos4<\/sup> \u03b8 + sin4<\/sup> \u03b8
= 2 cos2<\/sup> \u03b8 \u2013 (cos4<\/sup> \u03b8 \u2013 sin4<\/sup> \u03b8)
= 2 cos2<\/sup> \u03b8 \u2013 [(cos2<\/sup> \u03b8)2<\/sup> \u2013 (sin2<\/sup> \u03b8)2<\/sup>]
Using identity, (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)
= 2 cos2<\/sup> \u03b8 \u2013 [(cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8)(cos2<\/sup> \u03b8+ sin2<\/sup> \u03b8 )]
= 2 cos2<\/sup> \u03b8 \u2013 [(cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8)(1)] [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=2 cos2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8
= cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8
= 1 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 K <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

1 \u2013 2 cos2<\/sup> \u03b8 + cos4<\/sup> \u03b8 = sin4<\/sup>\u03b8<\/strong>
Sol :
Taking LHS = 1 \u2013 2 cos2<\/sup> \u03b8 + cos4<\/sup> \u03b8
We know that,
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
= 1\u2013 2 cos2<\/sup> \u03b8 + (cos2<\/sup> \u03b8)2<\/sup>
= 1 \u2013 2 cos2<\/sup> \u03b8 + (1 \u2013 sin2<\/sup> \u03b8)2<\/sup>
= 1 \u2013 2 cos2<\/sup> \u03b8 +1 + sin4<\/sup> \u03b8 \u2013 2sin2<\/sup>\u03b8
= 2 \u2013 2(cos2<\/sup> \u03b8 + sin2<\/sup>\u03b8) + sin4<\/sup> \u03b8
= 2 \u2013 2(1) + sin4<\/sup> \u03b8
= sin4<\/sup> \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 15 L <\/h4>\n\n\n\n

Prove the following identities :<\/strong><\/p>\n\n\n\n

1 \u2013 2 sin2<\/sup> \u03b8 + sin4<\/sup> \u03b8 = cos4<\/sup>\u03b8<\/strong>
Sol :
Taking LHS = 1 \u2013 2 sin2<\/sup> \u03b8 + sin4<\/sup> \u03b8
We know that,
cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1
= 1\u2013 2 sin2<\/sup> \u03b8 + (sin2<\/sup> \u03b8)2<\/sup>
= 1 \u2013 2 sin2<\/sup> \u03b8 + (1 \u2013 cos2<\/sup> \u03b8)2<\/sup>
= 1 \u2013 2 sin2<\/sup> \u03b8 +1 + cos4<\/sup> \u03b8 \u2013 2cos2<\/sup>\u03b8
= 2 \u2013 2(cos2<\/sup> \u03b8 + sin2<\/sup>\u03b8) + cos4<\/sup> \u03b8
= 2 \u2013 2(1) + cos4<\/sup> \u03b8
= cos4<\/sup> \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 16 A <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

sec2<\/sup>\u03b8 + cosec2<\/sup>\u03b8 = sec2<\/sup>\u03b8.cosec2<\/sup>\u03b8<\/strong>
Sol :
Taking LHS = sec2<\/sup> \u03b8 + cosec2<\/sup> \u03b8<\/p>\n\n\n\n

$=\\frac{1}{\\cos ^{2} \\theta}+\\frac{1}{\\sin ^{2} \\theta}$ $\\because\\left[\\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$ and $\\left[\\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
$=\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\cos ^{2} \\theta \\sin ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1}{\\cos ^{2} \\theta \\sin ^{2} \\theta}$
= sec2<\/sup> \u03b8 \u00d7 cosec2<\/sup> \u03b8 $\\because\\left[\\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$ and $\\left[\\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 16 B <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{\\cos ^{2} \\theta}{\\sin \\theta}+\\sin \\theta=\\operatorname{cosec} \\theta$
Sol :
Taking LHS $=\\frac{\\cos ^{2} \\theta}{\\sin \\theta}+\\sin \\theta$
$=\\frac{\\cos ^{2} \\theta+\\sin ^{2} \\theta}{\\sin \\theta}$
$=\\frac{1}{\\sin \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= cosec \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 16 C <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

cot\u03b8 + tan \u03b8 = cosec \u03b8 . sec \u03b8<\/strong>
Sol :
Taking LHS = cot \u03b8 + tan \u03b8<\/p>\n\n\n\n

$=\\frac{\\cos \\theta}{\\sin \\theta}+\\frac{\\sin \\theta}{\\cos \\theta}$<\/p>\n\n\n\n

$\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right]$ and $\\left[\\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
$=\\frac{\\cos ^{2} \\theta+\\sin ^{2} \\theta}{\\cos \\theta \\sin \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1}{\\cos \\theta \\sin \\theta}$
= cosec \u03b8 sec \u03b8 $\\because\\left[\\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$ and $\\left[\\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 17 <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{1-\\sin \\theta}{1+\\sin \\theta}=\\left(\\frac{1-\\sin \\theta}{\\cos \\theta}\\right)^{2}$
Sol :
Taking LHS $=\\frac{1-\\sin \\theta}{1+\\sin \\theta}$<\/p>\n\n\n\n

Multiplying and divide by the conjugate of 1 + sin \u03b8 , we get
$=\\frac{1-\\sin \\theta}{1+\\sin \\theta} \\times \\frac{1-\\sin \\theta}{1-\\sin \\theta}$
$=\\frac{(1-\\sin \\theta)^{2}}{(1)^{2}-(\\sin \\theta)^{2}}$
[\u2235 (a \u2013 b) (a \u2013 b) =(a \u2013 b)2<\/sup> and (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
$=\\frac{(1-\\sin \\theta)^{2}}{1-\\sin ^{2} \\theta}$
$=\\frac{(1-\\sin \\theta)^{2}}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\left(\\frac{1-\\sin \\theta}{\\cos \\theta}\\right)^{2}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 18 <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{1-\\cos \\theta}{1+\\cos \\theta}=\\left(\\frac{1-\\cos \\theta}{\\sin \\theta}\\right)^{2}$
Sol :
Taking LHS $=\\frac{1-\\cos \\theta}{1+\\cos \\theta}$<\/p>\n\n\n\n

Multiplying and divide by the conjugate of 1 + cos \u03b8 , we get
$=\\frac{1-\\cos \\theta}{1+\\cos \\theta} \\times \\frac{1-\\cos \\theta}{1-\\cos \\theta}$
$=\\frac{(1-\\cos \\theta)^{2}}{(1)^{2}-(\\cos \\theta)^{2}}$
[\u2235 (a \u2013 b) (a \u2013 b) =(a \u2013 b)2<\/sup> and (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
$=\\frac{(1-\\cos \\theta)^{2}}{1-\\cos ^{2} \\theta}$
$=\\frac{(1-\\cos \\theta)^{2}}{\\sin ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\left(\\frac{1-\\cos \\theta}{\\sin \\theta}\\right)^{2}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 19 <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{1-\\cos \\theta}{1+\\cos \\theta}=\\left(\\frac{1-\\cos \\theta}{\\sin \\theta}\\right)^{2}$
Sol :
Taking LHS $=\\frac{1-\\cos \\theta}{1+\\cos \\theta}$<\/p>\n\n\n\n

Multiplying and divide by the conjugate of 1 + cos \u03b8 , we get
$=\\frac{1-\\cos \\theta}{1+\\cos \\theta} \\times \\frac{1-\\cos \\theta}{1-\\cos \\theta}$
$=\\frac{(1-\\cos \\theta)^{2}}{(1)^{2}-(\\cos \\theta)^{2}}$
[\u2235 (a \u2013 b) (a \u2013 b) =(a \u2013 b)2<\/sup> and (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
$=\\frac{(1-\\cos \\theta)^{2}}{1-\\cos ^{2} \\theta}$
$=\\frac{(1-\\cos \\theta)^{2}}{\\sin ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\left(\\frac{1-\\cos \\theta}{\\sin \\theta}\\right)^{2}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 20 <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

$\\frac{\\cos \\theta}{1+\\sin \\theta}=\\frac{1-\\sin \\theta}{\\cos \\theta}$
Sol :
Taking LHS $=\\frac{\\cos \\theta}{1+\\sin \\theta}$<\/p>\n\n\n\n

Multiplying and divide by the conjugate of 1 + sin \u03b8 , we get
$=\\frac{\\cos \\theta}{1+\\sin \\theta} \\times \\frac{1-\\sin \\theta}{1-\\sin \\theta}$
$=\\frac{\\cos \\theta(1-\\sin \\theta)}{(1)^{2}-(\\sin \\theta)^{2}}$ [\u2235 (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
$=\\frac{\\cos \\theta(1-\\sin \\theta)}{1-\\sin ^{2} \\theta}$
$=\\frac{\\cos \\theta(1-\\sin \\theta)}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1-\\sin \\theta}{\\cos \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 21 <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

(sin8<\/sup>\u03b8 \u2013 cos8<\/sup>\u03b8) = (sin2<\/sup>\u03b8 \u2013 cos2<\/sup>\u03b8)(1 \u2013 2sin2<\/sup>\u03b8 .cos2<\/sup>\u03b8)<\/strong>
Sol :
Taking LHS
= sin8<\/sup> \u03b8 \u2013 cos8<\/sup> \u03b8
= (sin4<\/sup> \u03b8)2<\/sup> \u2013 (cos4<\/sup> \u03b8)2<\/sup>
= (sin4<\/sup> \u03b8 \u2013 cos4<\/sup> \u03b8)(sin4<\/sup> \u03b8+ cos4<\/sup> \u03b8 )
[\u2235 (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)]
= {(sin2<\/sup> \u03b8)2<\/sup> \u2013 (cos2<\/sup> \u03b8)2<\/sup>}{(sin2<\/sup> \u03b8)2<\/sup> + (cos2<\/sup> \u03b8)2<\/sup>}
= (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) (sin2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8) [(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) \u2013 2 sin2<\/sup> \u03b8 cos2<\/sup> \u03b8]
[ \u2235 (a2<\/sup> + b2<\/sup>) = (a +b)2<\/sup> \u2013 2ab]
= (1)[ sin2<\/sup> \u03b8 \u2013cos2<\/sup> \u03b8][(1) \u2013 2 sin2<\/sup> \u03b8 cos2<\/sup> \u03b8]
= (sin2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8)(1 \u2013 2 sin2<\/sup> \u03b8 cos2<\/sup> \u03b8)
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 22 <\/h4>\n\n\n\n

Prove that the following identities :<\/strong><\/p>\n\n\n\n

2(sin6<\/sup> \u03b8 \u2013 cos6<\/sup> \u03b8) \u2013 3(sin4<\/sup> \u03b8 + cos4<\/sup> \u03b8 ) + (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8)<\/strong>
Sol :
Taking LHS
= 2(sin6<\/sup> \u03b8 \u2013 cos6<\/sup> \u03b8) \u2013 3(sin4<\/sup> \u03b8+ cos4<\/sup> \u03b8 ) + (sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8)
= 2[(sin2<\/sup> \u03b8)3<\/sup> \u2013 (cos2<\/sup> \u03b8)3<\/sup> ] \u2013 3[(sin2<\/sup> \u03b8)2<\/sup> + (cos2<\/sup> \u03b8)2<\/sup> ]+1 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
Now, we use these identities, (a3<\/sup> \u2013 b3<\/sup>)= (a + b)3<\/sup> \u2013 3ab(a+b) and (a2<\/sup> + b2<\/sup>) = (a +b)2<\/sup> \u2013 2ab]
= 2[(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8)3<\/sup> \u2013 3sin2<\/sup>\u03b8 cos2<\/sup>\u03b8 (sin2<\/sup> \u03b8+ cos2<\/sup> \u03b8)] \u20133[(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) \u2013 2 sin2<\/sup> \u03b8 cos2<\/sup> \u03b8]+ 1
=2[(1) \u2013 3sin2<\/sup>\u03b8 cos2<\/sup>\u03b8 (1)] \u2013 3[(1) \u2013 2 sin2<\/sup> \u03b8 cos2<\/sup> \u03b8] + 1 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=2(1 \u2013 3 sin2<\/sup> \u03b8 cos2<\/sup> \u03b8 )\u2013 3 + 6sin2<\/sup> \u03b8 cos2<\/sup> \u03b8+ 1
= 2\u2013 6sin2<\/sup>\u03b8 cos2<\/sup>\u03b8 \u2013 2 + 6sin2<\/sup> \u03b8 cos2<\/sup> \u03b8
=0
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 23 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{\\cos A}{1-\\tan A}+\\frac{\\sin A}{1-\\cot A}=\\sin A+\\cos A$
Sol :
Taking LHS $=\\frac{\\cos A}{1-\\tan A}+\\frac{\\sin A}{1-\\cot A}$<\/p>\n\n\n\n

$=\\frac{\\cos A}{1-\\frac{\\sin A}{\\cos A}}+\\frac{\\sin A}{1-\\frac{\\cos A}{\\sin A}}$<\/p>\n\n\n\n

$\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right.$ and $\\left.\\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right]$
$=\\frac{\\cos A}{\\frac{\\cos A-\\sin A}{\\cos A}}+\\frac{\\sin A}{\\frac{\\sin A-\\cos A}{\\sin A}}$
$=\\frac{\\cos ^{2} A}{\\cos A-\\sin A}+\\frac{\\sin ^{2} A}{\\sin A-\\cos A}$
$=\\frac{\\cos ^{2} A-\\sin ^{2} A}{\\cos A-\\sin A}$<\/p>\n\n\n\n

Using the identity, (a2<\/sup> \u2013 b2<\/sup>)= (a + b) (a \u2013 b)
$=\\frac{(\\cos A-\\sin A)(\\cos A+\\sin A)}{(\\cos A-\\sin A)}$
= sin A +cos A
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 24 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{\\sin \\theta}{1+\\cos \\theta}+\\frac{1+\\cos \\theta}{\\sin \\theta}=\\frac{2}{\\sin \\theta}$
Sol :
Taking LHS $=\\frac{\\sin \\theta}{1+\\cos \\theta}+\\frac{1+\\cos \\theta}{\\sin \\theta}$
$=\\frac{\\sin ^{2} \\theta+(1+\\cos \\theta)^{2}}{(1+\\cos \\theta)(\\sin \\theta)}$
$=\\frac{\\sin ^{2} \\theta+1+\\cos ^{2} \\theta+2 \\cos \\theta}{(1+\\cos \\theta)(\\sin \\theta)}$
$=\\frac{2+2 \\cos \\theta}{(1+\\operatorname{ses} \\theta)(\\sin \\theta)}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{2(1+\\cos \\theta)}{(1+\\cos \\theta)(\\sin \\theta)}$
$=\\frac{2}{\\sin \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 25 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{1}{1+\\sin \\theta}+\\frac{1}{1-\\sin \\theta}=2 \\sec ^{2} \\theta$
Sol :
Taking LHS $=\\frac{1}{1+\\sin \\theta}+\\frac{1}{1-\\sin \\theta}$
$=\\frac{1-\\sin \\theta+1+\\sin \\theta}{(1+\\sin \\theta)(1-\\sin \\theta)}$<\/p>\n\n\n\n

Using the identity, (a2<\/sup> \u2013 b2<\/sup>)= (a + b) (a \u2013 b)
$=\\frac{2}{(1)^{2}-(\\sin \\theta)^{2}}$
$=\\frac{2}{1-\\sin ^{2} \\theta}$
$=\\frac{2}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= 2 sec2<\/sup> \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 26 <\/h4>\n\n\n\n

Prove the following identities<\/strong>
$\\frac{1+\\sin \\theta}{\\cos \\theta}+\\frac{\\cos \\theta}{1+\\sin \\theta}=2 \\sec \\theta$
Sol :
Taking LHS $=\\frac{1+\\sin \\theta}{\\cos \\theta}+\\frac{\\cos \\theta}{1+\\sin \\theta}$
$=\\frac{(1+\\sin \\theta)^{2}+\\cos ^{2} \\theta}{(1+\\sin \\theta)(\\cos \\theta)}$
$=\\frac{1+\\sin ^{2} \\theta+2 \\sin \\theta+\\cos ^{2} \\theta}{(\\cos \\theta)(1+\\sin \\theta)}$
$=\\frac{2+2 \\sin \\theta}{(\\cos \\theta)(1+\\sin \\theta)}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{2(1+\\sin \\theta)}{(\\cos \\theta)(1+\\sin \\theta)}$
$=\\frac{2}{\\cos \\theta}$
=2 sec \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 27 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{\\cos \\theta}{1-\\sin \\theta}+\\frac{\\cos \\theta}{1+\\sin \\theta}=\\frac{2}{\\cos \\theta}$
Sol :
Taking LHS $=\\frac{\\cos \\theta}{1-\\sin \\theta}+\\frac{\\cos \\theta}{1+\\sin \\theta}$
$=\\frac{\\cos \\theta(1+\\sin \\theta)+\\cos \\theta(1-\\sin \\theta)}{(1-\\sin \\theta)(1+\\sin \\theta)}$
$=\\frac{\\cos \\theta+\\cos \\theta \\sin \\theta+\\cos \\theta-\\cos \\theta \\sin \\theta}{(1-\\sin \\theta)(1+\\sin \\theta)}$
$=\\frac{2 \\cos \\theta}{(1-\\sin \\theta)(1+\\sin \\theta)}$<\/p>\n\n\n\n

Using the identity, (a2<\/sup> \u2013 b2<\/sup>)= (a + b) (a \u2013 b)
$=\\frac{2 \\cos \\theta}{(1)^{2}-(\\sin \\theta)^{2}}$
$=\\frac{2 \\cos \\theta}{1-\\sin ^{2} \\theta}$
$=\\frac{2 \\cos \\theta}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]<\/p>\n\n\n\n

$=\\frac{2}{\\cos \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 28 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{1}{1+\\cos \\theta}+\\frac{1}{1-\\cos \\theta}=\\frac{2}{\\sin ^{2} \\theta}$<\/strong>
Sol :
Taking LHS $=\\frac{1}{1+\\cos \\theta}+\\frac{1}{1-\\cos \\theta}$
$=\\frac{1-\\cos \\theta+1+\\cos \\theta}{(1+\\cos \\theta)(1-\\cos \\theta)}$<\/p>\n\n\n\n

Using the identity, (a2<\/sup> \u2013 b2<\/sup>)= (a + b) (a \u2013 b)
$=\\frac{2}{(1)^{2}-(\\cos \\theta)^{2}}$
$=\\frac{2}{1-\\cos ^{2} \\theta}$
$=\\frac{2}{\\sin ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 29 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{1}{1-\\sin \\theta}-\\frac{1}{1+\\sin \\theta}=\\frac{2 \\tan \\theta}{\\cos \\theta}$
Sol :
Taking LHS $=\\frac{1}{1-\\sin \\theta}-\\frac{1}{1+\\sin \\theta}$
$=\\frac{1+\\sin \\theta-1+\\sin \\theta}{(1+\\sin \\theta)(1-\\sin \\theta)}$<\/p>\n\n\n\n

Using the identity, (a2<\/sup> \u2013 b2<\/sup>)= (a + b) (a \u2013 b)
$=\\frac{2 \\sin \\theta}{(1)^{2}-(\\sin \\theta)^{2}}$
$=\\frac{2 \\sin \\theta}{1-\\sin ^{2} \\theta}$
$=\\frac{2 \\sin \\theta}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]<\/p>\n\n\n\n

$=\\frac{2 \\tan \\theta}{\\cos \\theta}$
$\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 30 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

cot2<\/sup>\u03b8 \u2013 cos2<\/sup>\u03b8 = cot2<\/sup>\u03b8 . cos2<\/sup>\u03b8<\/strong>
Sol :
Taking LHS = cot2<\/sup> \u03b8 \u2013 cos2<\/sup> \u03b8<\/p>\n\n\n\n

$=\\frac{\\cos ^{2} \\theta}{\\sin ^{2} \\theta}-\\cos ^{2} \\theta$ $\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right]$
$=\\frac{\\cos ^{2} \\theta-\\sin ^{2} \\theta \\cos ^{2} \\theta}{\\sin ^{2} \\theta}$
$=\\frac{\\cos ^{2} \\theta\\left(1-\\sin ^{2} \\theta\\right)}{\\sin ^{2} \\theta}$
 [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{\\cos ^{2} \\theta \\cos ^{2} \\theta}{\\sin ^{2} \\theta}$
= cot2<\/sup> \u03b8 cos2<\/sup> \u03b8 $\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 31 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

tan2<\/sup> \u03c6 \u2013 sin2<\/sup> \u03c6 \u2013 tan2<\/sup> \u03c6 . sin2<\/sup> \u03c6 = 0<\/strong>
Sol :
Taking LHS = tan2<\/sup> \u03c6 \u2013 sin2<\/sup> \u03c6 \u2013 tan2<\/sup> \u03c6 sin2<\/sup> \u03c6
$=\\frac{\\sin ^{2} \\varphi}{\\cos ^{2} \\varphi}-\\sin ^{2} \\varphi-\\frac{\\sin ^{2} \\varphi}{\\cos ^{2} \\varphi} \\sin ^{2} \\varphi$
$=\\frac{\\sin ^{2} \\varphi-\\sin ^{2} \\varphi \\cos ^{2} \\varphi-\\sin ^{4} \\varphi}{\\cos ^{2} \\varphi}$
$=\\frac{\\sin ^{2} \\varphi\\left(1-\\cos ^{2} \\varphi-\\sin ^{2} \\varphi\\right)}{\\cos ^{2} \\varphi}$
$=\\frac{\\sin ^{2} \\varphi\\left\\{1-\\left(\\cos ^{2} \\varphi+\\sin ^{2} \\varphi\\right)\\right]}{\\cos ^{2} \\varphi}$ [\u2235 cos2<\/sup> \u03c6 + sin2<\/sup> \u03c6 = 1]
$=\\frac{\\sin ^{2} \\varphi\\{1-1\\}}{\\cos ^{2} \\varphi}$
= 0
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 32 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

tan2<\/sup> \u03c6 + cot2<\/sup> \u03c6 + 2 = sec2<\/sup>\u03d5. cosec2<\/sup>\u03d5<\/strong>
Sol :
Taking LHS = tan2<\/sup> \u03c6 + cot2<\/sup> \u03c6 + 2
$=\\frac{\\sin ^{2} \\varphi}{\\cos ^{2} \\varphi}+\\frac{\\cos ^{2} \\varphi}{\\sin ^{2} \\varphi}+2$
$=\\frac{\\sin ^{4} \\varphi+\\cos ^{4} \\varphi+2 \\sin ^{2} \\varphi \\cos ^{2} \\varphi}{\\cos ^{2} \\varphi \\sin ^{2} \\varphi}$ [\u2235 (a + b)2<\/sup> = (a2<\/sup> + b2<\/sup> + 2ab)]
$=\\frac{\\left(\\sin ^{2} \\varphi+\\cos ^{2} \\varphi\\right)^{2}}{\\cos ^{2} \\varphi \\sin ^{2} \\varphi}$ [\u2235 cos2<\/sup> \u03c6 + sin2<\/sup> \u03c6 = 1]
$=\\frac{1}{\\cos ^{2} \\varphi \\sin ^{2} \\varphi}$
= sec2<\/sup> \u03c6 cosec2<\/sup> \u03c6 $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right.$ and $\\left.\\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 33 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{\\operatorname{cosec} \\theta+\\cot \\theta-1}{\\cot \\theta-\\operatorname{cosec} \\theta+1}=\\frac{1+\\cos \\theta}{\\sin \\theta}$<\/strong>
Sol :
Taking LHS $=\\frac{\\operatorname{cosec} \\theta+\\cot \\theta-1}{\\cot \\theta-\\operatorname{cosec} \\theta+1}$
$=\\frac{(\\cot \\theta+\\operatorname{cosec} \\theta)-\\left(\\operatorname{cosec}^{2} \\theta-\\cot ^{2} \\theta\\right)}{\\cot \\theta-\\operatorname{cosec} \\theta+1}$ [\u2235 cot2<\/sup> \u03b8 \u2013 cosec2<\/sup> \u03b8 = 1]
$=\\frac{(\\cot \\theta+\\operatorname{cosec} \\theta)-\\{(\\operatorname{cosec} \\theta+\\cot \\theta)(\\operatorname{cosec} \\theta-\\cot \\theta)\\}}{\\cot \\theta-\\operatorname{cosec} \\theta+1}$
$=\\frac{(\\cot \\theta+\\operatorname{cosec} \\theta)(1-\\operatorname{cosec} \\theta+\\cot \\theta)}{\\cot \\theta-\\operatorname{cosec} \\theta+1}$
= cot \u03b8 + cosec \u03b8<\/p>\n\n\n\n

$=\\frac{\\cos \\theta}{\\sin \\theta}+\\frac{1}{\\sin \\theta}$ $\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right.$ and $\\left.\\sin \\theta=\\frac{1}{\\csc \\theta}\\right]$
$=\\frac{1+\\cos \\theta}{\\sin \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 34 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{\\tan \\theta}{1-\\cot \\theta}+\\frac{\\cot \\theta}{1-\\tan \\theta}=1+\\tan \\theta+\\cot \\theta$
Sol :
Taking LHS $=\\frac{\\tan \\theta}{1-\\cot \\theta}+\\frac{\\cot \\theta}{1-\\tan \\theta}$<\/p>\n\n\n\n

$=\\frac{\\frac{\\sin \\theta}{\\cos \\theta}}{1-\\frac{\\cos \\theta}{\\sin \\theta}}+\\frac{\\frac{\\cos \\theta}{\\sin \\theta}}{1-\\frac{\\sin \\theta}{\\cos \\theta}}$ $\\left[\\because \\cot \\theta=\\frac{\\cos \\theta}{\\sin \\theta}\\right.$ and $\\left.\\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
$=\\frac{\\frac{\\sin \\theta}{\\sin \\theta-\\cos \\theta}}{\\sin \\theta}+\\frac{\\frac{\\cos \\theta}{\\sin \\theta}}{\\frac{\\cos \\theta-\\sin \\theta}{\\cos \\theta}}$
$=\\frac{\\sin \\theta}{\\cos \\theta} \\times \\frac{\\sin \\theta}{\\sin \\theta-\\cos \\theta}+\\frac{\\cos \\theta}{\\sin \\theta} \\times \\frac{\\cos \\theta}{\\cos \\theta-\\sin \\theta}$
$=\\frac{\\sin ^{2} \\theta}{\\cos \\theta(\\sin \\theta-\\cos \\theta)}+\\frac{\\cos ^{2} \\theta}{\\sin \\theta(\\cos \\theta-\\sin \\theta)}$
$=\\frac{\\sin ^{2} \\theta}{\\cos \\theta(\\sin \\theta-\\cos \\theta)}+\\frac{\\cos ^{2} \\theta}{\\sin \\theta\\{-(\\sin \\theta-\\cos \\theta)\\}}$
$=\\frac{\\sin ^{2} \\theta}{\\cos \\theta(\\sin \\theta-\\cos \\theta)}-\\frac{\\cos ^{2} \\theta}{\\sin \\theta(\\sin \\theta-\\cos \\theta)}$
$=\\frac{\\sin ^{3} \\theta-\\cos ^{3} \\theta}{(\\cos \\theta \\sin \\theta)(\\sin \\theta-\\cos \\theta)}$
$=\\frac{(\\sin \\theta-\\cos \\theta)\\left(\\sin ^{2} \\theta+\\cos ^{2} \\theta+\\sin \\theta \\cos \\theta\\right)}{(\\cos \\theta \\sin \\theta)(\\sin \\theta-\\cos \\theta)}$ [\u2235 (a3<\/sup> \u2013 b3<\/sup>)= (a \u2013 b)(a2<\/sup> +b2<\/sup> +ab)]
$=\\frac{1+\\sin \\theta \\cos \\theta}{\\cos \\theta \\sin \\theta}$
$=\\frac{1}{\\cos \\theta \\sin \\theta}+\\frac{\\sin \\theta \\cos \\theta}{\\cos \\theta \\sin \\theta}$
= tan \u03b8 cot \u03b8 + 1
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 35 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\frac{1-\\cos \\theta}{1+\\cos \\theta}=(\\cot \\theta-\\operatorname{cosec} \\theta)^{2}$<\/strong>
Sol :
Taking LHS $=\\frac{1-\\cos \\theta}{1+\\cos \\theta}$<\/p>\n\n\n\n

Multiplying and divide by the conjugate of 1 + cos \u03b8 , we get
$=\\frac{1-\\cos \\theta}{1+\\cos \\theta} \\times \\frac{1-\\cos \\theta}{1-\\cos \\theta}$
$=\\frac{(1-\\cos \\theta)^{2}}{(1)^{2}-(\\cos \\theta)^{2}}$
[\u2235 (a \u2013 b) (a \u2013 b) =(a \u2013 b)2<\/sup> and (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)]
$=\\frac{(1-\\cos \\theta)^{2}}{1-\\cos ^{2} \\theta}$
$=\\frac{(1-\\cos \\theta)^{2}}{\\sin ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\left(\\frac{1-\\cos \\theta}{\\sin \\theta}\\right)^{2}$
$=\\left(\\frac{1}{\\sin \\theta}-\\frac{\\cos \\theta}{\\sin \\theta}\\right)^{2}$
= (cosec \u03b8 \u2013 cot \u03b8)2<\/sup>
= { \u2013 (cot \u03b8 \u2013 cosec \u03b8)}2<\/sup>
= (cot \u03b8 \u2013 cosec \u03b8)2<\/sup>
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 36 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\sqrt{\\frac{1+\\cos \\theta}{1-\\cos \\theta}}=\\frac{1+\\cos \\theta}{\\sin \\theta}$<\/strong>
Sol :
Taking LHS $=\\sqrt{\\frac{1+\\cos \\theta}{1-\\cos \\theta}}$
$=\\sqrt{\\frac{1+\\cos \\theta}{1-\\cos \\theta} \\times \\frac{1+\\cos \\theta}{1+\\cos \\theta}}$ [multiplying and divide by conjugate of 1\u2013 cos\u03b8]
$=\\sqrt{\\frac{(1+\\cos \\theta)^{2}}{(1)^{2}-(\\cos \\theta)^{2}}}$
$=\\sqrt{\\frac{(1+\\cos \\theta)^{2}}{\\left(1-\\cos ^{2} \\theta\\right)}}$
$=\\sqrt{\\frac{(1+\\cos \\theta)^{2}}{\\sin ^{2} \\theta}}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1+\\cos \\theta}{\\sin \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 37 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\sqrt{\\frac{1+\\cos \\theta}{1-\\cos \\theta}}=\\frac{\\sin \\theta}{1-\\cos \\theta}$
Sol :
Taking LHS $=\\sqrt{\\frac{1+\\cos \\theta}{1-\\cos \\theta}}$
$=\\sqrt{\\frac{1+\\cos \\theta}{1-\\cos \\theta} \\times \\frac{1+\\cos \\theta}{1+\\cos \\theta}}$ [multiplying and divide by conjugate of 1\u2013 cos\u03b8]
$=\\sqrt{\\frac{(1+\\cos \\theta)^{2}}{(1)^{2}-(\\cos \\theta)^{2}}}$
$=\\sqrt{\\frac{(1+\\cos \\theta)^{2}}{\\left(1-\\cos ^{2} \\theta\\right)}}$
$=\\sqrt{\\frac{(1+\\cos \\theta)^{2}}{\\sin ^{2} \\theta}}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
$=\\frac{1+\\cos \\theta}{\\sin \\theta}$<\/p>\n\n\n\n

Multiply and divide by conjugate of 1+ cos\u03b8, we get
$=\\frac{1+\\cos \\theta}{\\sin \\theta} \\times \\frac{1-\\cos \\theta}{1-\\cos \\theta}$
$=\\frac{1-\\cos ^{2} \\theta}{\\sin \\theta \\times(1-\\cos \\theta)}$
$=\\frac{\\sin ^{2} \\theta}{\\sin \\theta \\times(1-\\cos \\theta)}$
$=\\frac{\\sin \\theta}{1-\\cos \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 38 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta}}=\\sec \\theta-\\tan \\theta$
Sol :
Taking LHS $=\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta}}$
$=\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta} \\times \\frac{1-\\sin \\theta}{1-\\sin \\theta}}$ [multiplying and divide by conjugate of 1+sin\u03b8]
$=\\sqrt{\\frac{(1-\\sin \\theta)^{2}}{(1)^{2}-(\\sin \\theta)^{2}}}$
$=\\sqrt{\\frac{(1-\\sin \\theta)^{2}}{\\left(1-\\sin ^{2} \\theta\\right)}}$
$=\\sqrt{\\frac{(1-\\sin \\theta)^{2}}{\\cos ^{2} \\theta}}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=$=\\frac{1-\\sin \\theta}{\\cos \\theta}$
$=\\frac{1}{\\cos \\theta}-\\frac{\\sin \\theta}{\\cos \\theta}$
= sec \u03b8 \u2013 tan \u03b8 <\/p>\n\n\n\n

$\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right.$ and $\\left.\\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 39 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta}}=\\frac{\\cos \\theta}{1+\\sin \\theta}$<\/strong>
Sol :
Taking LHS $=\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta} \\times \\frac{1+\\sin \\theta}{1+\\sin \\theta}}$ [multiplying and divide by of 1+ sin \u03b8]<\/p>\n\n\n\n

$=\\sqrt{\\frac{(1)^{2}-(\\sin \\theta)^{2}}{(1+\\sin \\theta)^{2}}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{1-\\sin ^{2} \\theta}{(1+\\sin \\theta)^{2}}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{\\cos ^{2} \\theta}{(1+\\sin \\theta)^{2}}}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]<\/p>\n\n\n\n

$=\\frac{\\cos \\theta}{1+\\sin \\theta}$
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 40 <\/h4>\n\n\n\n

Prove the following identities<\/strong><\/p>\n\n\n\n

$\\sqrt{\\frac{1+\\sin \\theta}{1-\\sin \\theta}}+\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta}}=2 \\sec \\theta$<\/strong>
Sol :
Taking LHS $=\\sqrt{\\frac{1+\\sin \\theta}{1-\\sin \\theta}}+\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta}}$<\/p>\n\n\n

[multiplying and divide by conjugate of 1\u2013 sin\u03b8 in 1st<\/sup> term and 1+sin\u03b8 in 2nd<\/sup> term]<\/p>\n\n\n\n

$=\\sqrt{\\frac{1+\\sin \\theta}{1-\\sin \\theta} \\times \\frac{1+\\sin \\theta}{1+\\sin \\theta}}+\\sqrt{\\frac{1-\\sin \\theta}{1+\\sin \\theta} \\times \\frac{1-\\sin \\theta}{1-\\sin \\theta}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{(1+\\sin \\theta)^{2}}{(1)^{2}-(\\sin \\theta)^{2}}}+\\sqrt{\\frac{(1-\\sin \\theta)^{2}}{(1)^{2}-(\\sin \\theta)^{2}}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{(1+\\sin \\theta)^{2}}{\\left(1-\\sin ^{2} \\theta\\right)}}+\\sqrt{\\frac{(1-\\sin \\theta)^{2}}{\\left(1-\\sin ^{2} \\theta\\right)}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{(1+\\sin \\theta)^{2}}{\\cos ^{2} \\theta}}+\\sqrt{\\frac{(1-\\sin \\theta)^{2}}{\\cos ^{2} \\theta}}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]<\/p>\n\n\n\n

=$\\frac{1+\\sin \\theta}{\\cos \\theta}+\\frac{1-\\sin \\theta}{\\cos \\theta}$<\/p>\n\n\n\n

$=\\frac{1+\\sin \\theta+1-\\sin \\theta}{\\cos \\theta}$<\/p>\n\n\n\n

$=\\frac{2}{\\cos \\theta}$
= 2 sec \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 41 <\/h4>\n\n\n\n

If sec\u03b8 + tan\u03b8=m and sec \u03b8 \u2013 tan\u03b8 = n, then prove that <\/strong>$\\sqrt{m n}=1$<\/strong><\/p>\n\n\n\n

Sol :
Given : sec\u03b8+tan\u03b8=m and sec \u03b8\u2013tan\u03b8=n
To Prove : \u221amn = 1
Taking LHS = \u221amn
Putting the value of m and n, we get
$=\\sqrt{(\\sec \\theta+\\tan \\theta)(\\sec \\theta-\\tan \\theta)}$
Using the identity, (a + b) (a \u2013 b) = (a2<\/sup> \u2013 b2<\/sup>)<\/p>\n\n\n\n

$=\\sqrt{\\sec ^{2} \\theta-\\tan ^{2} \\theta}$<\/p>\n\n\n\n

=\u221a(1) [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]
=\u00b11
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 42 <\/h4>\n\n\n\n

If cos\u03b8 + sin\u03b8 = 1, then prove that cos\u03b8 \u2013 sin \u03b8 = \u00b1 1.<\/strong><\/p>\n\n\n\n

Sol :
Given: cos \u03b8+sin\u03b8=1
On squaring both the sides, we get
(cos \u03b8 +sin \u03b8)2<\/sup> =(1)2<\/sup>
\u21d2 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 + 2sin\u03b8 cos \u03b8 = 1
\u21d2 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 \u2013 2sin\u03b8 cos \u03b8
[\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
\u21d2 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = (cos\u03b8 \u2013 sin\u03b8)2<\/sup>
[\u2235 (a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)]
\u21d2 1 = (cos \u03b8 \u2013 sin \u03b8)2<\/sup>
\u21d2 (cos \u03b8 \u2013 sin \u03b8) = \u00b11
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 43 <\/h4>\n\n\n\n

If sin\u03b8 + sin2<\/sup>\u03b8 = 1, then prove that cos2<\/sup>\u03b8 +1 cos4<\/sup>\u03b8 = 1<\/strong><\/p>\n\n\n\n

Sol :
Given : sin \u03b8 + sin2<\/sup> \u03b8 = 1
\u21d2 sin \u03b8 = 1 \u2013 sin2<\/sup> \u03b8
Taking LHS = cos2<\/sup>\u03b8+ cos4<\/sup>\u03b8
= cos2<\/sup> \u03b8 + (cos2<\/sup> \u03b8)2<\/sup>
= (1\u2013 sin2<\/sup> \u03b8) + (1\u2013 sin2<\/sup> \u03b8)2<\/sup> \u2026(i)
Putting sin \u03b8 = 1 \u2013 sin2<\/sup> \u03b8 in Eq. (i), we get
= sin \u03b8 + (sin \u03b8)2<\/sup>
= sin \u03b8 + sin2<\/sup> \u03b8
= 1 [Given: sin \u03b8 + sin2<\/sup> \u03b8 = 1]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 44 <\/h4>\n\n\n\n

If tan\u03b8 + sec\u03b8 = x, show that sin\u03b8 =<\/strong>$\\frac{x^{2}-1}{x^{2}+1}$<\/strong><\/p>\n\n\n\n

Sol :
To show : $\\sin \\theta=\\frac{x^{2}-1}{x^{2}+1}$
Taking RHS $=\\frac{x^{2}-1}{x^{2}+1}$
Given tan\u03b8+sec\u03b8=x<\/p>\n\n\n\n

$=\\frac{(\\tan \\theta+\\sec \\theta)^{2}-1}{(\\tan \\theta+\\sec \\theta)^{2}+1}$<\/p>\n\n\n\n

$=\\frac{\\tan ^{2} \\theta+\\sec ^{2} \\theta+2 \\tan \\theta \\sec \\theta-1}{\\tan ^{2} \\theta+\\sec ^{2} \\theta+2 \\tan \\theta \\sec \\theta+1}$<\/p>\n\n\n\n

$=\\frac{\\tan ^{2} \\theta+\\tan ^{2} \\theta+2 \\tan \\theta \\sec \\theta}{\\sec ^{2} \\theta-1+\\sec ^{2} \\theta+2 \\tan \\theta \\sec \\theta+1}$ [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]<\/p>\n\n\n\n

$=\\frac{2 \\tan ^{2} \\theta+2 \\tan \\theta \\sec \\theta}{2 \\sec ^{2} \\theta+2 \\tan \\theta \\sec \\theta}$<\/p>\n\n\n\n

$=\\frac{\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}+\\frac{\\sin \\theta}{\\cos \\theta} \\times \\frac{1}{\\cos \\theta}}{\\frac{1}{\\cos ^{2} \\theta}+\\frac{\\sin \\theta}{\\cos \\theta} \\times \\frac{1}{\\cos \\theta}}$ $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta} \\text { and } \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$<\/p>\n\n\n\n

$=\\frac{\\sin ^{2} \\theta+\\sin \\theta}{1+\\sin \\theta}$<\/p>\n\n\n\n

$=\\frac{\\sin \\theta(\\sin \\theta+1)}{1+\\sin \\theta}$
=sin \u03b8
=LHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 45 <\/h4>\n\n\n\n

If sin\u03b8 + cos\u03b8 = p and sec\u03b8 + cosec\u03b8 = q, then show q(p2<\/sup>\u20131) =2p<\/strong><\/p>\n\n\n\n

Sol :
Given: sin \u03b8 + cos \u03b8 = p and sec \u03b8 + cosec \u03b8 = q
To show q(p2<\/sup> \u2013 1) = 2p
Taking LHS = q(p2<\/sup> \u2013 1)
Putting the value of sin \u03b8 + cos \u03b8 = p and sec \u03b8 + cosec \u03b8 = q, we get
=(sec \u03b8 + cosec \u03b8){( sin \u03b8 + cos \u03b8 )2<\/sup> \u2013 1)
=(sec \u03b8 + cosec \u03b8){(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 + 2sin \u03b8 cos\u03b8) \u2013 1)}
[\u2235 (a + b)2<\/sup> = (a2<\/sup> + b2<\/sup> + 2ab)]
=(sec \u03b8 + cosec \u03b8)(1+2sin \u03b8 cos \u03b8 \u2013 1)
=(sec \u03b8 + cosec \u03b8)(2sin \u03b8cos\u03b8)<\/p>\n\n\n\n

$=\\left(\\frac{1}{\\cos \\theta}+\\frac{1}{\\sin \\theta}\\right) \\times(2 \\sin \\theta \\cos \\theta)$ $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta} \\text { and } \\sin \\theta=\\frac{1}{\\cos \\theta}\\right]$<\/p>\n\n\n\n

$=\\frac{\\sin \\theta+\\cos \\theta}{\\cos \\theta \\sin \\theta} \\times 2 \\sin \\theta \\cos \\theta$<\/p>\n\n\n\n

= 2(sin \u03b8 +cos \u03b8)
= 2p [ given sin \u03b8 + cos \u03b8 = p]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 46 <\/h4>\n\n\n\n

If x cos\u03b8 =a and y = a tan\u03b8, then prove that x2<\/sup>\u2013y2<\/sup>=a2<\/sup><\/strong><\/p>\n\n\n\n

Sol :
Given: x cos\u03b8 = a and y = a tan\u03b8<\/p>\n\n\n\n

\u21d2$\\mathrm{x}=\\frac{\\mathrm{a}}{\\cos \\theta}$ and $y=a \\tan \\theta$<\/p>\n\n\n\n

To Prove : x2<\/sup>\u2013y2<\/sup>=a2<\/sup>
Taking LHS = x2<\/sup>\u2013y2<\/sup>
Putting the values of x and y, we get<\/p>\n\n\n\n

$=\\left(\\frac{a}{\\cos \\theta}\\right)^{2}-(a \\tan \\theta)^{2}$<\/p>\n\n\n\n

$=\\frac{a^{2}}{\\cos ^{2} \\theta}-a^{2} \\tan ^{2} \\theta$<\/p>\n\n\n\n

$=\\frac{a^{2}}{\\cos ^{2} \\theta}-a^{2} \\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}$ $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$<\/p>\n\n\n\n

$=\\frac{a^{2}-a^{2} \\sin ^{2} \\theta}{\\cos ^{2} \\theta}$<\/p>\n\n\n\n

$=\\frac{a^{2}\\left(1-\\sin ^{2} \\theta\\right)}{\\cos ^{2} \\theta}$<\/p>\n\n\n\n

$=\\frac{a^{2} \\cos ^{2} \\theta}{\\cos ^{2} \\theta}$ [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
= a2<\/sup>
= RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 47 <\/h4>\n\n\n\n

If x= r cos \u03b1 sin \u03b2, y = r sin \u03b1 sin \u03b2 and z = r cos \u03b1 then prove that x2<\/sup> + y2<\/sup> + z2<\/sup> = r2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :
Taking LHS = x2<\/sup> + y2<\/sup> + z2<\/sup>
Putting the values of x, y and z , we get
=(r cos \u03b1 sin \u03b2)2<\/sup> + (r sin \u03b1 sin \u03b2)2<\/sup> + (r cos \u03b1)2<\/sup>
= r2<\/sup> cos2<\/sup>\u03b1 sin2<\/sup>\u03b2 + r2<\/sup> sin2<\/sup>\u03b1 sin2<\/sup>\u03b2 + r2<\/sup> cos2<\/sup>\u03b1
Taking common r2<\/sup> sin2<\/sup> \u03b1 , we get
= r2<\/sup> sin2<\/sup>\u03b1 (cos2<\/sup>\u03b2 + sin2<\/sup> \u03b2) + r2<\/sup>cos2<\/sup> \u03b1
= r2<\/sup> sin2<\/sup>\u03b1 + r2<\/sup> cos2<\/sup> \u03b1 [\u2235 cos2<\/sup> \u03b2 + sin2<\/sup> \u03b2 = 1]
=r2<\/sup> ( sin2<\/sup> \u03b1 + cos2<\/sup> \u03b1)
= r2<\/sup> [\u2235 cos2<\/sup> \u03b1 + sin2<\/sup> \u03b1 = 1]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 48 <\/h4>\n\n\n\n

If sec\u03b8 \u2013 tan\u03b8 = x, then prove that<\/strong><\/p>\n\n\n\n

(i)<\/strong>$\\cos \\theta=\\frac{2 \\mathrm{x}}{1+\\mathrm{x}^{2}}$<\/strong>
(ii) <\/strong>$\\sin \\theta=\\frac{1-x^{2}}{1+x^{2}}$<\/strong>
Sol :
(i) Given sec \u03b8 \u2013 tan \u03b8 = x
Taking RHS $=\\frac{2 \\mathrm{x}}{1+\\mathrm{x}^{2}}$
Putting the value of x, we get
$=\\frac{2(\\sec \\theta-\\tan \\theta)}{1+(\\sec \\theta-\\tan \\theta)^{2}}$<\/p>\n\n\n\n

$=\\frac{2(\\sec \\theta-\\tan \\theta)}{1+\\sec ^{2} \\theta+\\tan ^{2} \\theta-2 \\sec \\theta \\tan \\theta}$<\/p>\n\n\n\n

$=\\frac{2(\\sec \\theta-\\tan \\theta)}{\\sec ^{2} \\theta+\\sec ^{2} \\theta-2 \\sec \\theta \\tan \\theta}$ [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]<\/p>\n\n\n\n

$=\\frac{2(\\sec \\theta-\\tan \\theta)}{2 \\sec \\theta(\\sec \\theta-\\tan \\theta)}$<\/p>\n\n\n\n

$=\\frac{1}{\\sec \\theta}$ $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right]$<\/p>\n\n\n\n

= cos \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

(ii) Given sec \u03b8 \u2013 tan \u03b8 = x<\/p>\n\n\n\n

Taking RHS $=\\frac{1-x^{2}}{1+x^{2}}$<\/p>\n\n\n\n

Putting the value of x, we get<\/p>\n\n\n\n

$=\\frac{1-(\\sec \\theta-\\tan \\theta)^{2}}{1+(\\sec \\theta-\\tan \\theta)^{2}}$<\/p>\n\n\n\n

$=\\frac{1-\\sec ^{2} \\theta-\\tan ^{2} \\theta+2 \\sec \\theta \\tan \\theta}{1+\\sec ^{2} \\theta+\\tan ^{2} \\theta-2 \\sec \\theta \\tan \\theta}$<\/p>\n\n\n\n

$=\\frac{\\left.-\\tan ^{2} \\theta-\\tan ^{2} \\theta+2 \\sec \\theta \\tan \\theta\\right)}{\\sec ^{2} \\theta+\\sec ^{2} \\theta-2 \\sec \\theta \\tan \\theta}$ [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]<\/p>\n\n\n\n

$=\\frac{2 \\tan \\theta(\\sec \\theta-\\tan \\theta)}{2 \\sec \\theta(\\sec \\theta-\\tan \\theta)}$<\/p>\n\n\n\n

=$\\frac{\\sin \\theta}{\\cos \\theta} \\times \\cos \\theta$ $\\left[\\because \\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}\\right]$
= sin \u03b8
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 49 <\/h4>\n\n\n\n

If a cos \u03b8 + b sin\u03b8 = c, then prove that a sin\u03b8 \u2013 b cos \u03b8 = \u00b1  <\/strong>$\\sqrt{a^{2}+b^{2}-c^{2}}$<\/strong><\/p>\n\n\n\n

Sol :
Let
(a cos \u03b8 + b sin \u03b8)2<\/sup> + (a sin \u03b8 \u2013 b cos \u03b8)2<\/sup> = a2<\/sup>cos2<\/sup>\u03b8 + b2<\/sup> sin2<\/sup>\u03b8 + 2ab cos \u03b8 sin \u03b8 + a2<\/sup>sin2<\/sup>\u03b8
+ b2<\/sup> cos2<\/sup>\u03b8 \u2013 2ab cos \u03b8 sin \u03b8
\u21d2 c2<\/sup> + (a sin \u03b8 \u2013 b cos \u03b8)2<\/sup> = a2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) + b2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8)
\u21d2 c2<\/sup> + (a sin \u03b8 \u2013 b cos \u03b8)2<\/sup> = a2<\/sup> + b2<\/sup>
\u21d2 (a sin \u03b8 \u2013 b cos \u03b8)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 c2<\/sup>
\u21d2 (a sin \u03b8 \u2013 b cos \u03b8) = \u00b1\u221a (a2<\/sup> + b2<\/sup> \u2013 c2<\/sup>)<\/p>\n\n\n\n

\n\n\n\n

Question 50 <\/h4>\n\n\n\n

If 1+sin2<\/sup>\u03b8 = 3 sin\u03b8 . cos\u03b8, then prove that tan \u03b8 = 1 or 1\/2.<\/strong><\/p>\n\n\n\n

Sol :
Given: 1+sin2<\/sup>\u03b8 =3 sin \u03b8 cos \u03b8
Divide by cos2<\/sup> \u03b8 to both the sides, we get<\/p>\n\n\n\n

\u21d2$\\frac{1}{\\cos ^{2} \\theta}+\\frac{\\sin ^{2} \\theta}{\\cos ^{2} \\theta}=\\frac{3 \\sin \\theta \\cos \\theta}{\\cos ^{2} \\theta}$<\/p>\n\n\n\n

\u21d2 sec2<\/sup> \u03b8 + tan2<\/sup> \u03b8 = 3 tan \u03b8
\u21d2 1+ tan2<\/sup> \u03b8+ tan2<\/sup> \u03b8 = 3tan \u03b8
\u21d2 2 tan2<\/sup> \u03b8 \u20133tan \u03b8 +1 = 0
Let tan\u03b8 = x
\u21d2 2x2<\/sup> \u2013 3x +1 = 0
\u21d2 2x2<\/sup> \u2013 2x \u2013 x +1 = 0
\u21d2 2x ( x \u2013 1) \u2013 1(x \u2013 1) = 0
\u21d2 (2x \u2013 1)(x \u2013 1) = 0
Putting each of the factor = 0, we get
\u21d2 x = 1 or $\\dfrac{1}{2}$
And above, we let tan \u03b8 = x<\/p>\n\n\n\n

\u21d2$\\tan \\theta=1$ or $\\frac{1}{2}$<\/p>\n\n\n\n

Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 51 <\/h4>\n\n\n\n

If a cos \u03b8 \u2013 b sin\u03b8 = x and a sin\u03b8 + b cos\u03b8 = y that a2<\/sup> + b2<\/sup> = x2<\/sup> + y2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :
Taking RHS =x2<\/sup> + y2<\/sup>
Putting the values of x and y, we get
(a cos \u03b8 \u2013 b sin \u03b8)2<\/sup> + (a sin \u03b8 + b cos \u03b8)2<\/sup>
= a2<\/sup>cos2<\/sup>\u03b8 + b2<\/sup> sin2<\/sup>\u03b8 \u2013 2ab cos \u03b8 sin \u03b8 + a2<\/sup>sin2<\/sup>\u03b8 + b2<\/sup> cos2<\/sup>\u03b8 + 2ab cos \u03b8 sin \u03b8
= a2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8) + b2<\/sup> (cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8)
= a2<\/sup> + b2<\/sup> [\u2235 cos2<\/sup> \u03b8 + sin2<\/sup> \u03b8 = 1]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 52 <\/h4>\n\n\n\n

If x = a sec \u03b8 + b tan \u03b8 a and y = a tan \u03b8 + b sec \u03b8, then prove that x2<\/sup> \u2013 y2<\/sup> = a2<\/sup> \u2013 b2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :
Taking LHS =x2<\/sup> \u2013 y2<\/sup>
Putting the values of x and y, we get
(a sec \u03b8 + b tan \u03b8)2<\/sup> \u2013 (a tan \u03b8 + b sec \u03b8)2<\/sup>
= a2<\/sup> sec2<\/sup>\u03b8 + b2<\/sup> tan2<\/sup>\u03b8 + 2ab sec \u03b8 tan \u03b8 \u2013 a2<\/sup>tan2<\/sup>\u03b8 \u2013 b2<\/sup> sec2<\/sup>\u03b8 \u2013 2ab sec \u03b8 tan \u03b8
= a2<\/sup> (sec2<\/sup>\u03b8 \u2013 tan2<\/sup>\u03b8) \u2013 b2<\/sup> (sec2<\/sup>\u03b8 \u2013 tan2<\/sup>\u03b8)
= a2<\/sup> \u2013 b2<\/sup> [\u2235 1+ tan2<\/sup> \u03b8 = sec2<\/sup> \u03b8]
=RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 53 <\/h4>\n\n\n\n

If (a2<\/sup> \u2013 b2<\/sup>) sin \u03b8 + 2ab cos\u03b8 = a2<\/sup> + b2<\/sup>, then prove that .<\/strong><\/p>\n\n\n\n

$\\tan \\theta=\\frac{\\mathrm{a}^{2}-\\mathrm{b}^{2}}{2 \\mathrm{ab}}$
Sol :
Taking (a2<\/sup> \u2013 b2<\/sup>) sin \u03b8 + 2ab cos \u03b8 = a2<\/sup> + b2<\/sup>
We know that<\/p>\n\n\n\n

$\\sin \\theta=\\frac{2 \\tan \\frac{\\theta}{2}}{1+\\tan ^{2} \\frac{\\theta}{2}}$ and $\\cos \\theta=\\frac{1-\\tan ^{2} \\frac{\\theta}{2}}{1+\\tan ^{2} \\frac{\\theta}{2}}$<\/p>\n\n\n\n

Then, substituting the above values in the given equation, we get<\/p>\n\n\n\n

$=\\mathrm{a}^{2}-\\mathrm{b}^{2} \\frac{2 \\tan \\frac{\\theta}{2}}{1+\\tan ^{2} \\frac{\\theta}{2}}+2 \\mathrm{ab} \\frac{1-\\tan ^{2} \\frac{\\theta}{2}}{1+\\tan ^{2} \\frac{\\theta}{2}}=\\mathrm{a}^{2}+\\mathrm{b}^{2}$<\/p>\n\n\n\n

Now, substituting,$t=\\tan \\frac{\\theta}{2}$ , we have<\/p>\n\n\n\n

$a^{2}-b^{2} \\frac{2 t}{1+t^{2}}+2 a b \\frac{1-t^{2}}{1+t^{2}}=a^{2}+b^{2}$<\/p>\n\n\n\n

\u21d2 ( a2<\/sup> \u2013 b2<\/sup>)2t \u2013 2ab(1 \u2013 t2<\/sup>) = (a2<\/sup> + b2<\/sup>)(1+t2<\/sup>)
Simplify, we get
(a2<\/sup> + 2ab + b2<\/sup>)t2<\/sup> \u2013 2(a2<\/sup> \u2013 b2<\/sup>)t + (a2<\/sup> \u20132ab +b2<\/sup>)=0
\u21d2 (a+b)2<\/sup> t2<\/sup> \u2013 2(a2<\/sup> \u2013 b2<\/sup>)t + (a \u2013 b)2<\/sup> = 0
\u21d2 (a+b)2<\/sup> t2<\/sup> \u20132 (a \u2013 b)(a+b)t + (a \u2013 b)2<\/sup> =0
\u21d2 [(a+b)t \u2013 (a \u2013 b)]2<\/sup> = 0 [\u2235 (a \u2013 b)2<\/sup> = (a2<\/sup> + b2<\/sup> \u2013 2ab)]
\u21d2 [(a+b)t \u2013 (a \u2013 b)] = 0
\u21d2 (a+b)t = (a \u2013 b)
\u21d2$ \\mathrm{t}=\\frac{\\mathrm{a}-\\mathrm{b}}{\\mathrm{a}+\\mathrm{b}}$<\/p>\n\n\n\n

\u21d2$ \\tan \\frac{\\theta}{2}=\\frac{a-b}{a+b}$<\/p>\n\n\n\n

We know that, $\\frac{2 t}{1-t^{2}}=\\tan \\theta$ where $\\mathrm{t}=\\tan \\frac{\\theta}{2}$<\/p>\n\n\n\n

$\\Rightarrow \\tan \\theta=\\frac{2\\left(\\frac{\\mathrm{a}-\\mathrm{b}}{\\mathrm{a}+\\mathrm{b}}\\right)}{1-\\left(\\frac{\\mathrm{a}-\\mathrm{b}}{\\mathrm{a}+\\mathrm{b}}\\right)^{2}}$<\/p>\n\n\n\n

$\\Rightarrow \\tan \\theta=\\frac{2(\\mathrm{a}+\\mathrm{b})(\\mathrm{a}-\\mathrm{b})}{(\\mathrm{a}+\\mathrm{b})^{2}-(\\mathrm{a}-\\mathrm{b})^{2}}$<\/p>\n\n\n\n

$\\Rightarrow \\tan \\theta=\\frac{a^{2}-b^{2}}{2 a b}$<\/p>\n\n\n\n

Hence Proved<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 Fill in the blanks (i) sin2 \u03b8 cosec2 \u03b8 = \u2026\u2026..(ii) 1 + tan2 \u03b8 = \u2026\u2026(iii) Reciprocal sin \u03b8. cot \u03b8 = \u2026\u2026(iv) 1\u2013…….= cos2\u03b8(v) $\\tan A=\\frac{\\cdots \\cdots \\cdot}{\\cos A}$(vi) $\\ldots \\ldots=\\frac{\\cos A}{\\sin A}$(vii) cos \u03b8 is reciprocal of ………(viii) Reciprocal of sin \u03b8 is………(ix) Value of sin \u03b8 in terms of cos \u03b8 is(x) Value of cos \u03b8 in terms of sin \u03b8 isSol […]<\/p>\n","protected":false},"author":302,"featured_media":624083,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-624101","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 4.4 - 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