{"id":624094,"date":"2023-09-01T06:37:57","date_gmt":"2023-09-01T06:37:57","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624094"},"modified":"2023-09-04T10:19:08","modified_gmt":"2023-09-04T10:19:08","slug":"kc-sinha-exercise-4-2-mathematics-solution-class-10-chapter-4-trigonometric-ratios-and-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-4-2-mathematics-solution-class-10-chapter-4-trigonometric-ratios-and-identities\/","title":{"rendered":"KC Sinha: Exercise 4.2 – Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities"},"content":{"rendered":"\n\n\n\n\n

Question 1 <\/h4>\n\n\n\n

Find the value of the following :<\/strong><\/p>\n\n\n\n

(i) sin 30o<\/sup> + cos 60o<\/sup><\/strong>
(ii) sin2<\/sup> 45o<\/sup>+cos2<\/sup>45o<\/sup><\/strong>
(iii) sin 30o<\/sup> + cos 60o<\/sup> \u2013 tan45o<\/sup><\/strong>
(iv) <\/strong>$\\sqrt{1+\\tan ^{2} 60^{\\circ}}$<\/strong>
(v) tan 60o<\/sup> x cos30o<\/sup><\/strong>
Sol :
(i) sin 30o<\/sup> + cos 60o<\/sup>
We know that,<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}>\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

So,
sin(30o<\/sup>) + cos(60o<\/sup>)
$=\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{2}\\right)$
=1<\/p>\n\n\n\n

(ii) sin2<\/sup> 45o<\/sup>+cos2<\/sup>45o<\/sup>
We know that,
$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

So, sin2<\/sup> 45o<\/sup>+cos2<\/sup>45o<\/sup>
$=\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}+\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}$
$=\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{2}\\right)$
=1<\/p>\n\n\n\n

(iii) sin 30o<\/sup> + cos 60o<\/sup> \u2013 tan45o<\/sup>
$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$
$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$
tan(45o<\/sup>)=1<\/p>\n\n\n\n

So,
sin 30o<\/sup> + cos 60o<\/sup> \u2013 tan 45o<\/sup>
$=\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{2}\\right)-1$
$=\\frac{1+1-2}{2}$
=0<\/p>\n\n\n\n

(iv) $\\sqrt{1+\\tan ^{2} 60^{\\circ}}$
We know that
tan(60o<\/sup>) = \u221a3
So,
$=\\sqrt{1+\\tan ^{2} 60^{\\circ}}$
$=\\sqrt{1+(\\sqrt{3})^{2}}$
$=\\sqrt{1+3}$
=\u221a4
= 2
(v) tan 60o<\/sup> \u00d7 cos30o<\/sup>
tan(60o<\/sup>) = \u221a3
$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$
So,
tan 60o<\/sup> \u00d7 cos30o<\/sup>
$=\\sqrt{3} \\times \\frac{\\sqrt{3}}{2}$
$=\\frac{3}{2}$<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

If \u03b8 = 45\u00b0, find the value of<\/strong><\/p>\n\n\n\n

(i) <\/strong>$\\tan ^{2} \\theta+\\frac{1}{\\sin ^{2} \\theta}$<\/strong>
(ii) cos2<\/sup> \u03b8 – sin2<\/sup> \u03b8<\/strong>
Sol :
(i) $\\tan ^{2} \\theta+\\frac{1}{\\sin ^{2} \\theta}$
Given \u03b8 =45\u00b0
We know that,
tan(45o<\/sup>) = 1
$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
$=(1)^{2}+\\frac{1}{\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}}$
= 1+ 2
= 3
(ii) cos2<\/sup> \u03b8 \u2013 sin2<\/sup> \u03b8
Given \u03b8 = 45\u00b0
$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
So, cos2<\/sup> 45\u00b0 \u2013 sin2<\/sup> 45\u00b0
$=\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}-\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}$
= 0<\/p>\n\n\n\n

\n\n\n\n

Question 3 A <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

sin45\u00b0.cos45\u00b0 \u2013 sin2<\/sup>30\u00b0.<\/strong>
Sol :
We know that,
$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Now, putting the values
$=\\left(\\frac{1}{\\sqrt{2}}\\right) \\times\\left(\\frac{1}{\\sqrt{2}}\\right)-\\left(\\frac{1}{2}\\right)^{2}$
$=\\left(\\frac{1}{2}\\right)-\\left(\\frac{1}{4}\\right)$
$=\\left(\\frac{1}{4}\\right)$<\/p>\n\n\n\n

\n\n\n\n

Question 3 B <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\frac{\\tan 60^{\\circ}}{\\sin 60^{\\circ}+\\cos 60^{\\circ}}$
Sol :
We know that,
$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$
$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$
tan (60\u00b0<\/sup>) = \u221a3<\/p>\n\n\n\n

Now putting the values;<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{\\frac{\\sqrt{3}}{2}+\\frac{1}{2}}$<\/p>\n\n\n\n

$=\\frac{\\frac{\\sqrt{3}}{1+\\sqrt{3}}}{2}$<\/p>\n\n\n\n

$=\\sqrt{3} \\times \\frac{2}{1+\\sqrt{3}}$<\/p>\n\n\n\n

$=\\frac{2 \\sqrt{3}}{1+\\sqrt{3}}$<\/p>\n\n\n\n

Multiplying and dividing by the conjugate of (1+\u221a3)
$=\\frac{2 \\sqrt{3}}{1+\\sqrt{3}} \\times \\frac{1-\\sqrt{3}}{1-\\sqrt{3}}$
$=\\frac{2 \\sqrt{3}-6}{(1)^{2}-(\\sqrt{3})^{2}}$ [\u2235(a)2<\/sup> \u2013 (b)2<\/sup> = (a+b)(a-b)]
$=\\frac{2 \\sqrt{3}-6}{-2}$<\/p>\n\n\n\n

Multiplying and dividing by (-2)
= 3 – \u221a3<\/p>\n\n\n\n

\n\n\n\n

Question 3 C <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\frac{\\tan 60^{\\circ}}{\\sin 60^{\\circ}+\\cos 30^{\\circ}}$
Sol :
We know that,
$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$
$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$
tan (60o<\/sup>) = \u221a3<\/p>\n\n\n\n

Now putting the values;<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{\\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{3}}{2}}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{\\sqrt{3}}$<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

\n\n\n\n

Question 3 D <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\frac{4}{\\sin ^{2} 60^{\\circ}}+\\frac{3}{\\cos ^{2} 60^{\\circ}}$
Sol :
We know that<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Now putting the value, we get
$=\\frac{4}{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}}+\\frac{3}{\\left(\\frac{1}{2}\\right)^{2}}$
$=4 \\times\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}+3 \\times(2)^{2}$
$=4\\left(\\frac{4}{3}\\right)+3 \\times 4$
$=\\frac{16}{3}+12$
$=\\frac{16+36}{3}$
$=\\frac{52}{3}$<\/p>\n\n\n\n

\n\n\n\n

Question 3 E <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

sin2<\/sup> 60\u00b0 \u2013 cos2<\/sup> 60\u00b0<\/strong>
Sol :
We know that,<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Now putting the value;
$=\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}$
$=\\left(\\frac{3}{4}\\right)-\\left(\\frac{1}{4}\\right)$
$=\\frac{1}{2}$<\/p>\n\n\n\n

\n\n\n\n

Question 3 F <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

4sin2<\/sup> 30\u00b0 + 3 tan 30\u00b0 \u2013 8 sin 45\u00b0 cos 45\u00b0<\/strong>
Sol :
We know that,<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Now putting the value, we get
$=4 \\times\\left(\\frac{1}{2}\\right)^{2}+3 \\times\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}-8 \\times \\frac{1}{\\sqrt{2}} \\times \\frac{1}{\\sqrt{2}}$
$=4 \\times \\frac{1}{4}+3 \\times \\frac{1}{3}-8 \\times \\frac{1}{2}$
= 1 + 1 \u2013 4
= -2<\/p>\n\n\n\n

\n\n\n\n

Question 3 G <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

2sin2<\/sup>30\u00b0 \u2013 3cos2<\/sup> 45\u00b0 + tan2<\/sup> 60\u05c4\u00b0<\/strong>
Sol :
We know that,<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Tan (60o<\/sup>) = \u221a3
Now putting the value;
$=2 \\times\\left(\\frac{1}{2}\\right)^{2}-3 \\times\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}+(\\sqrt{3})^{2}$
$=2 \\times \\frac{1}{4}-3 \\times \\frac{1}{2}+3$
$=\\frac{1}{2}-\\frac{3}{2}+3$<\/p>\n\n\n\n

$=\\frac{1-3+6}{2}$<\/p>\n\n\n\n

$=\\frac{4}{2}$<\/p>\n\n\n\n

=2<\/p>\n\n\n\n

\n\n\n\n

Question 3 H <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

sin 90\u00b0 + cos 0\u00b0 + sin 30\u00b0 + cos 60\u00b0<\/strong>
Sol :
We know that,
Sin (90o<\/sup>) = 1
Cos (0o<\/sup>) = 1<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Now putting the value;<\/p>\n\n\n\n

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

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

$=\\frac{6}{2}$<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

\n\n\n\n

Question 3 I <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

sin 90\u00b0 \u2013 cos 0\u00b0 + tan 0\u00b0 + tan 45\u00b0<\/strong>
Sol :
We know that
Sin (90o<\/sup>) = 1
Cos (0o<\/sup>) = 1
Tan(0o<\/sup>) = 0
Tan(45o<\/sup>) = 1
Now putting the value, we get
= 1 \u2013 1 + 0 + 1
= 1<\/p>\n\n\n\n

\n\n\n\n

Question 3 J <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\cos ^{2} 0^{\\circ}+\\tan ^{2} \\frac{\\pi}{4}+\\sin ^{2} \\frac{\\pi}{4}$, where \u03c0 = 180\u00b0<\/strong>
Sol :
We know that
Cos (0o<\/sup>) = 1
Tan (45o<\/sup>) = 1 $\\left[\\frac{\\pi}{4}=\\frac{180^{\\circ}}{4}=45^{\\circ}\\right]$
$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}\\left[\\frac{\\pi}{4}=\\frac{180^{\\circ}}{4}=45^{\\circ}\\right]$<\/p>\n\n\n\n

Now putting the values;
$=(1)^{2}+(1)^{2}+\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}$
$=1+1+\\frac{1}{2}$
$=\\frac{2+2+1}{2}$
$=\\frac{5}{2}$<\/p>\n\n\n\n

\n\n\n\n

Question 3 K <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\frac{\\cos 60^{\\circ}}{\\sin ^{2} 45^{\\circ}}-3 \\cot 45^{\\circ}+2 \\sin 90^{\\circ}$
Sol :
We know that,
$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$
$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$
Cot (45o<\/sup>) = 1
Sin (90o<\/sup>) = 1
Now putting the values, we get
$=\\frac{\\frac{1}{2}}{\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}}-3(1)+2(1)$
= 1-3+2
=0<\/p>\n\n\n\n

\n\n\n\n

Question 3 L <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\frac{4}{\\tan ^{2} 60^{\\circ}}+\\frac{1}{\\cos ^{2} 30^{\\circ}}-\\sin ^{2} 45^{\\circ}$
Sol :
We can write the above equation as:
= 4 cot2<\/sup> 60o<\/sup> + sec2<\/sup> 30o<\/sup> \u2013 sin2<\/sup> 45o<\/sup> \u2026(a) $\\left[\\because \\cos \\theta=\\frac{1}{\\sec \\theta}\\right.$ and $\\left.\\tan \\theta=\\frac{1}{\\cot \\theta}\\right]$<\/p>\n\n\n\n

$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\cot \\left(60^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

$\\operatorname{Sec}\\left(30^{\\circ}\\right)=\\frac{2}{\\sqrt{3}}$<\/p>\n\n\n\n

Now putting the values in (a);
$=4\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}+\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}-\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}$<\/p>\n\n\n\n

$=4 \\times \\frac{1}{3}+\\frac{4}{3}-\\frac{1}{2}$<\/p>\n\n\n\n

$=\\frac{8+8-3}{6}$<\/p>\n\n\n\n

$=\\frac{13}{6}$<\/p>\n\n\n\n

\n\n\n\n

Question 3 M <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

cos60\u00b0 . cos 30\u00b0 \u2013 sin 60\u00b0 . sin 30\u00b0<\/strong>
Sol :
We know that,<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Now putting the values, we get<\/p>\n\n\n\n

$=\\frac{1}{2} \\times \\frac{\\sqrt{3}}{2}-\\frac{\\sqrt{3}}{2} \\times \\frac{1}{2}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4}-\\frac{\\sqrt{3}}{4}$<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

\n\n\n\n

Question 3 N <\/h4>\n\n\n\n

Find the numerical value of the following :<\/strong><\/p>\n\n\n\n

$\\frac{4\\left(\\sin ^{2} 60^{\\circ}-\\cos ^{2} 45^{\\circ}\\right)}{\\tan ^{2} 30^{\\circ}+\\cos ^{2} 90^{\\circ}}$
Sol :
We know that,<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

cos(90o<\/sup>) = 0<\/p>\n\n\n\n

Now putting the values;
$=4 \\times \\frac{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}}{\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}-(0)^{2}}$
$=4 \\times \\frac{\\frac{3}{4}-\\frac{1}{2}}{\\frac{1}{3}}$
$=4 \\times \\frac{1}{4} \\times 3$
= 3<\/p>\n\n\n\n

\n\n\n\n

Question 4 A <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

sin30\u00b0.cos45\u00b0 + cos30\u00b0.sin45\u00b0<\/strong>
Sol :<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Now putting the values, we get<\/p>\n\n\n\n

$=\\frac{1}{2} \\times \\frac{1}{\\sqrt{2}}+\\frac{\\sqrt{3}}{2} \\times \\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$=\\frac{1}{2 \\sqrt{2}}+\\frac{\\sqrt{3}}{2 \\sqrt{2}}$<\/p>\n\n\n\n

$=\\frac{1+\\sqrt{3}}{2 \\sqrt{2}}$<\/p>\n\n\n\n

\n\n\n\n

Question 4 B <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

cosec2<\/sup>30\u00b0.tan2<\/sup>45\u00b0 \u2013 sec2<\/sup>60\u00b0<\/strong>
Sol :
We know that
cosec (30o<\/sup>) = 2
Tan(45o<\/sup>) = 1
sec (60 o<\/sup>) = 2
Now putting the values;
= (2)2<\/sup> \u00d7 (1)2<\/sup> – (2)2<\/sup>
= 4 \u2013 4
= 0<\/p>\n\n\n\n

\n\n\n\n

Question 4 C <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

2sin2<\/sup>30\u00b0.tan60\u00b0 \u2013 3cos2<\/sup>60\u00b0.sec2<\/sup>30\u00b0<\/strong>
Sol :
We know that
$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$
tan (60o<\/sup>) = \u221a3<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\sec \\left(30^{\\circ}\\right)=\\frac{2}{\\sqrt{3}}$<\/p>\n\n\n\n

Now putting the values;<\/p>\n\n\n\n

$=2 \\times\\left(\\frac{1}{2}\\right)^{2} \\times(\\sqrt{3})-\\left(3 \\times\\left(\\frac{1}{2}\\right)^{2} \\times\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}\\right)$<\/p>\n\n\n\n

$=2 \\times \\frac{1}{4} \\times(\\sqrt{3})-\\left(3 \\times \\frac{1}{4} \\times \\frac{4}{3}\\right)$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{2}-1$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}-2}{2}$<\/p>\n\n\n\n

\n\n\n\n

Question 4 D <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

tan60\u00b0 . cosec2<\/sup>45\u00b0 + sec2<\/sup>60\u00b0.tan45\u00b0<\/strong>
Sol :
We know that
tan (60o<\/sup>) = \u221a3
cosec (45o<\/sup>) = \u221a2
sec (60 o<\/sup>) = 2
tan(45o<\/sup>) = 1
Now putting the values;
= (\u221a3) \u00d7 (\u221a2)2<\/sup> + (2)2<\/sup> \u00d7 (1)
= 2\u221a3 +4
=2 (\u221a3 + 2)<\/p>\n\n\n\n

\n\n\n\n

Question 4 E <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

tan30\u00b0.sec45\u00b0 + tan60\u00b0.sin30\u00b0<\/strong>
Sol :
We know that
$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$
sec (45o<\/sup>) = \u221a2
tan (60o<\/sup>) = \u221a3
$\\sec \\left(30^{\\circ}\\right)=\\frac{2}{\\sqrt{3}}$<\/p>\n\n\n\n

Now putting the values, we get
$=\\frac{1}{\\sqrt{3}} \\times \\sqrt{2}+\\sqrt{3} \\times \\frac{2}{\\sqrt{3}}$
$=\\frac{\\sqrt{2}}{\\sqrt{3}}+2$
$=2+\\frac{\\sqrt{2}}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{\\sqrt{3}}$
$=2+\\frac{\\sqrt{6}}{3}$<\/p>\n\n\n\n

\n\n\n\n

Question 4 F <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

cos30\u00b0.cos45\u00b0 \u2013 sin30\u00b0.sin45\u00b0<\/strong>
Sol :
We know that<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Now putting the values, we get
$=\\frac{\\sqrt{3}}{2} \\times \\frac{1}{\\sqrt{2}}-\\frac{1}{2} \\times \\frac{1}{\\sqrt{2}}$
$=\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}$<\/p>\n\n\n\n

Multiplying and dividing by (\u221a2), we get<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}-1}{2 \\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{2}(\\sqrt{3}-1)}{4}$<\/p>\n\n\n\n

\n\n\n\n

Question 4 G <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

$\\frac{4}{3} \\tan ^{2} 30^{\\circ}+\\sin ^{2} 60^{\\circ}-3 \\cos ^{2} 60^{\\circ}+\\frac{3}{4} \\tan ^{2} 60^{\\circ}-2 \\tan ^{2} 45^{\\circ}$
Sol :
We know that
$\\begin{aligned} \\tan \\left(30^{\\circ}\\right) &=\\frac{1}{\\sqrt{3}} \\\\ \\sin \\left(60^{\\circ}\\right) &=\\frac{\\sqrt{3}}{2} \\\\ \\cos \\left(60^{\\circ}\\right) &=\\frac{1}{2} \\end{aligned}$
tan (60o<\/sup>) = \u221a3
tan(45o<\/sup>) = 1<\/p>\n\n\n\n

Now putting the values;
$=\\left(\\frac{4}{3} \\times\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\right)+\\left[\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}\\right]-\\left[3 \\times\\left(\\frac{1}{2}\\right)^{2}\\right]+\\left[\\frac{3}{4}(\\sqrt{3})^{2}\\right]-\\left[2 \\times(1)^{2}\\right]$
$=\\left[\\frac{4}{3} \\times \\frac{1}{3}\\right]+\\left[\\frac{3}{4}\\right]-\\left[3 \\times \\frac{1}{4}\\right]+\\left[\\frac{3}{4} \\times 3\\right]-[2 \\times(1)]$<\/p>\n\n\n\n

$=\\frac{4}{9}+\\frac{3}{4}-\\frac{3}{4}+\\frac{9}{4}-2$<\/p>\n\n\n\n

$=\\frac{16+27-27+81-72}{36}$<\/p>\n\n\n\n

$=\\frac{25}{36}$<\/p>\n\n\n\n

\n\n\n\n

Question 4 H <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

$\\frac{\\tan ^{2} 60^{\\circ}+4 \\cos ^{2} 45^{\\circ}+3 \\sec ^{2} 30^{\\circ}+5 \\cos ^{2} 90^{\\circ}}{\\operatorname{cosec} 30^{\\circ}+\\sec 60^{\\circ}-\\cot ^{2} 30^{\\circ}}$
Sol :
We know that
tan (60o<\/sup>) = \u221a3<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\sec \\left(30^{\\circ}\\right)=\\frac{2}{\\sqrt{3}}$<\/p>\n\n\n\n

cos(90o<\/sup>) = 0
cosec (30o<\/sup>) = 2
sec (60 o<\/sup>) = 2
cot (30o<\/sup>) = \u221a3<\/p>\n\n\n\n

Now putting the values, we get
$=\\frac{(\\sqrt{3})^{2}+\\left[4 \\times\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}\\right]+\\left[3 \\times\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}\\right]+\\left[5 \\times(0)^{2}\\right]}{(2)+(2)-(\\sqrt{3})^{2}}$
$=\\frac{(3)+\\left[4 \\times \\frac{1}{2}\\right]+\\left[3 \\times \\frac{4}{3}\\right]+[5 \\times 0]}{2+2-3}$
$=\\frac{(3)+[2]+[4]+[0]}{2+2-3}$
= 9<\/p>\n\n\n\n

\n\n\n\n

Question 4 I <\/h4>\n\n\n\n

Evaluate the following :<\/strong><\/p>\n\n\n\n

$\\frac{5 \\sin ^{2} 30^{\\circ}+\\cos ^{2} 45^{\\circ}-4 \\tan ^{2} 30^{\\circ}}{2 \\sin 30^{\\circ} \\cdot \\cos 30^{\\circ}+\\tan 45^{\\circ}}$
Sol :
We know that<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

tan (45o)<\/sup> = 1<\/p>\n\n\n\n

Now putting the values, we get
$=\\frac{\\left[5 \\times\\left(\\frac{1}{2}\\right)^{2}\\right]+\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}-\\left[4 \\times\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\right]}{2\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)+(1)}$
$=\\frac{\\left(\\frac{5}{4}\\right)+\\left(\\frac{1}{2}\\right)-\\left(\\frac{4}{3}\\right)}{\\left(\\frac{\\sqrt{3}}{2}\\right)+1}$
$=\\frac{\\left(\\frac{15+6-16}{12}\\right)}{\\left(\\frac{\\sqrt{3}+2}{2}\\right)}$
$=\\frac{5}{12} \\times \\frac{2}{\\sqrt{3}+1}$
$=\\frac{5}{6} \\times \\frac{1}{\\sqrt{3}+2}$<\/p>\n\n\n\n

\n\n\n\n

Question 5 A <\/h4>\n\n\n\n

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

$\\frac{(1-\\cos B)(1+\\cos B)}{(1-\\sin B)(1+\\sin B)}=\\frac{1}{3}$ When B = 30\u00b0<\/strong>
Sol :
Solving, L.H.S.
$=\\frac{(1)^{2}-(\\cos B)^{2}}{(1)^{2}-(\\sin B)^{2}}$ [(a)2<\/sup> \u2013 (b)2<\/sup> = (a+b)(a-b)]
$=\\frac{1-\\cos ^{2} B}{1-\\sin ^{2} B}$<\/p>\n\n\n\n

We know that,
$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$
$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Putting the values, we get
$=\\frac{1-\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}}{1-\\left(\\frac{1}{2}\\right)^{2}}$
$=\\frac{1-\\frac{3}{4}}{1-\\frac{1}{4}}$
$=\\frac{4-3}{4-1}$
$=\\frac{1}{3}$
=R.H.S.
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 5 B <\/h4>\n\n\n\n

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

$\\frac{(1-\\cos \\alpha)(1+\\cos \\alpha)}{(1-\\sin \\alpha)(1+\\sin \\alpha)}=3$ When \u03b1 =60\u00b0<\/strong>
Sol :
Solving, L.H.S.
$=\\frac{(1)^{2}-(\\cos \\alpha)^{2}}{(1)^{2}-(\\sin \\alpha)^{2}}$ [(a)2<\/sup> \u2013 (b)2<\/sup> = (a+b)(a-b)]
$=\\frac{1-\\cos ^{2} \\alpha}{1-\\sin ^{2} \\alpha}$<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

Putting the values, we get
$=\\frac{1-\\left(\\frac{1}{2}\\right)^{2}}{1-\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}}$<\/p>\n\n\n\n

$=\\frac{1-\\frac{1}{4}}{1-\\frac{3}{4}}$<\/p>\n\n\n\n

$=\\frac{4-1}{4-3}$
= 3 = R.H.S.<\/p>\n\n\n\n

\n\n\n\n

Question 5 C <\/h4>\n\n\n\n

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

cos(A \u2013 B) = cos A. cos B + sinA . sin B if A=B=60o<\/sup><\/strong>
Sol :
Solving, L.H.S.
= cos (60o<\/sup> \u2013 60o<\/sup>) [Putting the value A=B=60o<\/sup>]
= cos (0o<\/sup>)
= 1
Solving, R.H.S.
= cos (60o<\/sup>) \u00d7 cos (60o<\/sup>) + sin (60o<\/sup>) \u00d7 sin (60o<\/sup>) [Putting the value A=B=60o<\/sup>]
= cos2<\/sup>(60o<\/sup>) + sin2<\/sup>(60o<\/sup>)<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$=\\left(\\frac{1}{2}\\right)^{2}+\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}$<\/p>\n\n\n\n

$=\\frac{1}{4}+\\frac{3}{4}$<\/p>\n\n\n\n

$=\\frac{1+3}{4}$<\/p>\n\n\n\n

= 1
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 5 D <\/h4>\n\n\n\n

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

4(sin4<\/sup>30\u00b0 + cos4<\/sup> 60\u00b0) \u2013 3(cos2<\/sup> 45\u00b0 \u2013 sin2<\/sup>90\u00b0) = 2<\/strong>
Sol :
We know that,<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Sin (90o<\/sup>) = 1<\/p>\n\n\n\n

= 4[{(sin 30o<\/sup>)2<\/sup>}2<\/sup> + {(cos 60o<\/sup>)2<\/sup>}2<\/sup>] \u2013 3[(cos 45o<\/sup>)2<\/sup> – (sin 90o<\/sup>)2<\/sup>]<\/p>\n\n\n\n

Putting the values
$=4 \\times\\left[\\left\\{\\left(\\frac{1}{2}\\right)^{2}\\right\\}^{2}+\\left\\{\\left(\\frac{1}{2}\\right)^{2}\\right\\}^{2}\\right]-3\\left[\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}-1\\right]$
$=4 \\times\\left[\\left\\{\\frac{1}{4}\\right\\}^{2}+\\left\\{\\frac{1}{4}\\right\\}^{2}\\right]-3\\left[\\frac{1}{2}-1\\right]$
$=4 \\times\\left[\\frac{1}{16}+\\frac{1}{16}\\right]-3\\left[-\\frac{1}{2}\\right]$
$=4 \\times\\left[\\frac{1}{8}\\right]-3\\left[-\\frac{1}{2}\\right]$<\/p>\n\n\n\n

$=\\left[\\frac{1}{2}\\right]+\\left[\\frac{3}{2}\\right]$<\/p>\n\n\n\n

$=\\left[\\frac{4}{2}\\right]$<\/p>\n\n\n\n

=2 = R.H.S.
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 5 E <\/h4>\n\n\n\n

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

sin90\u00b0 = 2sin45\u00b0.cos45\u00b0<\/strong>
Sol :
We know that,
sin (90o<\/sup>) = 1<\/p>\n\n\n\n

$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Taking LHS = sin 90\u00b0 = 1<\/p>\n\n\n\n

Now, taking RHS<\/p>\n\n\n\n

$=2 \\times \\frac{1}{\\sqrt{2}} \\times \\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$=2 \\times \\frac{1}{2}$<\/p>\n\n\n\n

= 1
= R.H.S.
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 5 F <\/h4>\n\n\n\n

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

cos60\u00b0 = 2cos2<\/sup>30\u00b0 \u2013 1 = 1 \u2013 2 sin2<\/sup>30\u00b0<\/strong>
Sol :
We know that,<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Taking LHS = cos 60\u00b0 $=\\frac{1}{2}$<\/p>\n\n\n\n

Now, solving RHS = 2cos2<\/sup> 30\u00b0 – 1 , we get<\/p>\n\n\n\n

$=2 \\times\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-1$<\/p>\n\n\n\n

$=2 \\times \\frac{3}{4}-1$<\/p>\n\n\n\n

$=\\frac{3}{2}-1$<\/p>\n\n\n\n

$=\\frac{3}{2}-1$<\/p>\n\n\n\n

$=\\frac{1}{2}$<\/p>\n\n\n\n

= RHS<\/p>\n\n\n\n

Now taking RHS = 1- 2sin2<\/sup> 30\u00b0<\/p>\n\n\n\n

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

$=1-\\frac{1}{2}$<\/p>\n\n\n\n

$=\\frac{2-1}{2}$<\/p>\n\n\n\n

$=\\frac{1}{2}$<\/p>\n\n\n\n

= RHS
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 G <\/h4>\n\n\n\n

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

cos90\u00b0 = 1 \u2013 2 sin2<\/sup>45\u00b0 = 2cos2<\/sup>45\u00b0 \u2013 1<\/strong>
Sol :
We know that
cos(90o<\/sup>) = 0<\/p>\n\n\n\n

$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

taking LHS = cos 90\u00b0 = 0<\/p>\n\n\n\n

Now solving RHS 1- 2sin2<\/sup> 45\u00b0<\/p>\n\n\n\n

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

$=1-2 \\times \\frac{1}{2}$<\/p>\n\n\n\n

= 1- 1
= 0
= RHS<\/p>\n\n\n\n

Now, solving RHS = 2cos2<\/sup> 45\u00b0 – 1 , we get<\/p>\n\n\n\n

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

$=1-2 \\times \\frac{1}{2}$<\/p>\n\n\n\n

= 1- 1
= 0
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 H <\/h4>\n\n\n\n

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

sin30\u00b0.cos60\u00b0 + cos30\u00b0.sin60\u00b0 = sin90\u00b0<\/strong>
Sol :
We know that<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

sin (90o<\/sup>) = 1<\/p>\n\n\n\n

Taking LHS =
$=\\left[\\left(\\frac{1}{2}\\right) \\times\\left(\\frac{1}{2}\\right)\\right]+\\left[\\left(\\frac{\\sqrt{3}}{2}\\right) \\times\\left(\\frac{\\sqrt{3}}{2}\\right)\\right]$<\/p>\n\n\n\n

$=\\left[\\left(\\frac{1}{4}\\right)\\right]+\\left[\\left(\\frac{3}{4}\\right)\\right]$<\/p>\n\n\n\n

$=\\left[\\frac{1+3}{4}\\right]$<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

Now, RHS = sin 90\u00b0 = 1
\u2234 LHS = RHS
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 I <\/h4>\n\n\n\n

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

cos60\u00b0.cos30\u00b0 \u2013 sin60\u00b0. sin30\u00b0 = cos 90\u00b0<\/strong>
Sol :
We know that<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

cos(90o<\/sup>) = 0<\/p>\n\n\n\n

Taking LHS
$=\\left[\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\right]-\\left[\\left(\\frac{\\sqrt{3}}{2}\\right)\\left(\\frac{1}{2}\\right)\\right]$
$=\\left[\\left(\\frac{\\sqrt{3}}{4}\\right)\\right]-\\left[\\left(\\frac{\\sqrt{3}}{4}\\right)\\right]$
= 0<\/p>\n\n\n\n

\u2234 LHS =RHS
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 J <\/h4>\n\n\n\n

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

$\\cos 60^{\\circ}=\\frac{1-\\tan ^{2} 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}$
Sol :
We know that,<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

Taking LHS = $\\cos 60^{\\circ}=\\frac{1}{2}$<\/p>\n\n\n\n

Now, solving RHS
$=\\frac{1-\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}}{1+\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}}$
$=\\frac{1-\\frac{1}{3}}{1+\\frac{1}{3}}$
$=\\frac{\\frac{3-1}{3}}{\\frac{3+1}{3}}$<\/p>\n\n\n\n

$=\\frac{2}{4}$<\/p>\n\n\n\n

$=\\frac{1}{2}$<\/p>\n\n\n\n

\u2234 L.H.S. = R.H.S.
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 K <\/h4>\n\n\n\n

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

$\\frac{\\tan 60^{\\circ}-\\tan 30^{\\circ}}{1+\\tan 60^{\\circ} \\cdot \\tan 30^{\\circ}}=\\tan 30^{\\circ}$
Sol :
We know that
tan(60o<\/sup>) = \u221a3
$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

Taking LHS
$=\\frac{\\sqrt{3}-\\frac{1}{\\sqrt{3}}}{1+(\\sqrt{3}) \\times\\left(\\frac{1}{\\sqrt{3}}\\right)}$
$=\\frac{\\frac{3-1}{\\sqrt{3}}}{1+1}$
$=\\frac{\\frac{2}{\\sqrt{3}}}{2}$
$=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

Now, RHS $=\\tan 30^{\\circ}=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

\u2234L.H.S. = R.H.S.
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 L <\/h4>\n\n\n\n

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

$\\frac{1-\\tan 30^{\\circ}}{1+\\tan 30^{\\circ}}=\\frac{1-\\sin 60^{\\circ}}{\\cos 60^{\\circ}}$
Sol :<\/p>\n\n\n\n

$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Taking LHS
$=\\frac{1-\\frac{1}{\\sqrt{3}}}{1+\\frac{1}{\\sqrt{3}}}$
$=\\frac{\\frac{\\sqrt{3}-1}{\\sqrt{3}}}{\\frac{\\sqrt{3}+1}{\\sqrt{3}}}$
$=\\frac{\\sqrt{3}-1}{\\sqrt{3}+1}$<\/p>\n\n\n\n

Multiplying and Dividing, LHS by (\u221a3- 1)
$=\\frac{\\sqrt{3}-1}{\\sqrt{3}+1} \\times \\frac{\\sqrt{3}-1}{\\sqrt{3}-1}$
$=\\frac{(\\sqrt{3}-1)^{2}}{(\\sqrt{3})^{2}-(1)^{2}}$ [(a)2<\/sup> \u2013 (b)2<\/sup> = (a+b)(a-b)]
$=\\frac{(\\sqrt{3}-1)^{2}}{(\\sqrt{3})^{2}-(1)^{2}}$
$=\\frac{3+1-2 \\sqrt{3}}{3-1}$
$=\\frac{4-2 \\sqrt{3}}{2}$<\/p>\n\n\n\n

Multiplying and Dividing, LHS by 2
= 2- \u221a3<\/p>\n\n\n\n

Now, RHS
$=\\frac{1-\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}$
$=\\frac{\\frac{2-\\sqrt{3}}{2}}{\\frac{1}{2}}$
= 2- \u221a3<\/p>\n\n\n\n

\u2234 LHS = RHS
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

Question 5 M <\/h4>\n\n\n\n

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

$\\frac{\\sin 60^{\\circ}+\\cos 30^{\\circ}}{\\sin 30^{\\circ}+\\cos 60^{\\circ}+1}=\\cos 30^{\\circ}$
Sol :
We know that<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

Taking LHS
$=\\frac{\\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}+\\frac{1}{2}+1}$
$=\\frac{2 \\times \\frac{\\sqrt{3}}{2}}{\\frac{1+1+2}{2}}$
$=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

Now, RHS= $\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$
\u2234 LHS =RHS<\/p>\n\n\n\n

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

\n\n\n\n

Question 5 N <\/h4>\n\n\n\n

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

$\\sin 60^{\\circ}=2 \\sin 30^{\\circ} \\cdot \\cos 30^{\\circ}=\\frac{2 \\tan 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}$
Sol :
We know that<\/p>\n\n\n\n

$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$<\/p>\n\n\n\n

$\\sin \\left(60^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$\\tan \\left(30^{\\circ}\\right)=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

Taking LHS $=\\sin 60^{\\circ}=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

Now, solving RHS = 2 sin 30\u00b0 cos 30\u00b0<\/p>\n\n\n\n

$=2 \\times \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

= LHS<\/p>\n\n\n\n

Now, RHS$=\\frac{2 \\tan 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}$
$=\\frac{2\\left(\\frac{1}{\\sqrt{3}}\\right)}{1+\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}}$
$=\\frac{\\frac{2}{\\sqrt{3}}}{1+\\frac{1}{3}}$
$=\\frac{\\frac{2}{\\sqrt{3}}}{\\frac{4}{3}}$
$=\\frac{2}{\\sqrt{3}} \\times \\frac{3}{4}$
$=\\frac{\\sqrt{3}}{2}$
\u2234 LHS =RHS
Hence proved<\/p>\n\n\n\n

\n\n\n\n

Question 6A <\/h4>\n\n\n\n

If A=60o<\/sup> and B = 30o<\/sup>, verify that :<\/strong><\/p>\n\n\n\n

cos (A+B) = cos A cos B \u2013 sin A sin B<\/strong>
Sol :
Given: A=60o<\/sup> and B =30o<\/sup>
Now, LHS = Cos (A+B)
\u21d2 Cos (60 o<\/sup> + 30 o<\/sup>)
\u21d2 Cos (90 o<\/sup>)
\u21d2 0 [\u2235 cos 90 o<\/sup> = 0]
Now, RHS = Cos A Cos B \u2013 Sin A Sin B
\u21d2 cos(60 o<\/sup>) cos(30 o<\/sup>) \u2013 sin(60 o<\/sup>) sin (30 o<\/sup>)
$\\Rightarrow\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{3}}{2}\\right)\\left(\\frac{1}{2}\\right)$
\u21d2 0
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 6 B <\/h4>\n\n\n\n

If A=60o<\/sup> and B = 30o<\/sup>, verify that :<\/strong><\/p>\n\n\n\n

sin (A \u2013 B) = sin A cos B \u2013 cos A sin B<\/strong>
Sol :
Given: A=60o<\/sup> and B =30o<\/sup>
Now, LHS = Sin (A-B)
\u21d2 Sin (60 o<\/sup> – 30 o<\/sup>)
\u21d2 Sin (30 o<\/sup>)
$\\Rightarrow\\left(\\frac{1}{2}\\right)$<\/p>\n\n\n\n

Now, RHS = Sin A Cos B \u2013 Cos A Sin B<\/p>\n\n\n\n

\u21d2 sin(60 o<\/sup>) cos(30 o<\/sup>) \u2013 cos(60 o<\/sup>) sin (30 o<\/sup>)
$\\Rightarrow\\left(\\frac{\\sqrt{3}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{2}\\right)$
$\\Rightarrow \\frac{3}{4}-\\frac{1}{4}$
$\\Rightarrow\\left(\\frac{1}{2}\\right)$
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 6 C <\/h4>\n\n\n\n

If A=60o<\/sup> and B = 30o<\/sup>, verify that :<\/strong><\/p>\n\n\n\n

tan (A \u2013 B) <\/strong>$=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$<\/strong>
Sol :
Given: A=60o<\/sup> and B =30o<\/sup>
Now, LHS = tan (A-B)
\u21d2 tan (60 o<\/sup> – 30 o<\/sup>)
\u21d2 tan (30 o<\/sup>)
$\\Rightarrow\\left(\\frac{1}{\\sqrt{3}}\\right)$<\/p>\n\n\n\n

Now, RHS $=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$
$=\\frac{\\tan 60^{\\circ}-\\tan 30^{\\circ}}{1+\\tan 60^{\\circ} \\tan 30^{\\circ}}$
$\\Rightarrow \\frac{\\sqrt{3}-\\left(\\frac{1}{\\sqrt{3}}\\right)}{1+(\\sqrt{3})\\left(\\frac{1}{\\sqrt{3}}\\right)}$
$\\Rightarrow \\frac{\\frac{3-1}{\\sqrt{3}}}{1+1}$
$\\Rightarrow\\left(\\frac{1}{\\sqrt{3}}\\right)$
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 7 A <\/h4>\n\n\n\n

If A = 30o<\/sup>, verify that :<\/strong><\/p>\n\n\n\n

sin 2A = 2 sin A cos A<\/strong>
Sol :
Given: A =30o<\/sup>
Now, LHS = sin 2(30o<\/sup>)
\u21d2 sin 60o<\/sup>
$\\Rightarrow \\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

Now, RHS = 2 sin A cos A
\u21d2 2 sin (30o<\/sup>) cos (30o<\/sup>)
$\\Rightarrow 2\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)$
$\\Rightarrow \\frac{\\sqrt{3}}{2}$
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 7 B <\/h4>\n\n\n\n

If A = 30o<\/sup>, verify that :<\/strong><\/p>\n\n\n\n

cos 2A = 1-2 sin2<\/sup>A=2cos2<\/sup> A \u2013 1<\/strong>
Sol :
Given: A =30o<\/sup><\/p>\n\n\n\n

Now, LHS = cos 2(30o<\/sup>)
\u21d2 cos 60o<\/sup>
$\\Rightarrow \\frac{1}{2}$<\/p>\n\n\n\n

Now, RHS = 1- 2sin2<\/sup> A
\u21d2 1- 2sin2<\/sup> (30o<\/sup>)
$\\Rightarrow 1-2\\left(\\frac{1}{2}\\right)^{2}$
$\\Rightarrow 1-2\\left(\\frac{1}{4}\\right)$
$\\Rightarrow \\frac{2-1}{2}$
$\\Rightarrow \\frac{1}{2}$<\/p>\n\n\n\n

Now, RHS = 2cos2<\/sup> A \u2013 1
\u21d2 2cos2<\/sup> (30o<\/sup>) – 1
$\\Rightarrow 2\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-1$
$\\Rightarrow 2\\left(\\frac{3}{4}\\right)-1$
$\\Rightarrow \\frac{3-2}{2}$
$\\Rightarrow \\frac{1}{2}$<\/p>\n\n\n\n

\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 8 A <\/h4>\n\n\n\n

If \u03b8 = 30\u00b0, verify that :<\/strong>
sin 3\u03b8 = 3 sin\u03b8 \u2013 4 sin3<\/sup>\u03b8<\/strong>
Sol :
Given: \u03b8 =30o<\/sup>
Now, LHS = sin 3(30o<\/sup>)
\u21d2 sin 90o<\/sup>
= 1
Now, RHS = 3 sin \u03b8 – 4 sin3<\/sup> \u03b8
\u21d2 3 sin (30o<\/sup>) – 4 sin3<\/sup> (30o<\/sup>)
$\\Rightarrow 3\\left(\\frac{1}{2}\\right)-4\\left(\\frac{1}{2}\\right)^{3}$
$\\Rightarrow \\frac{3}{2}-\\frac{1}{2}$
= 1
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 8 B <\/h4>\n\n\n\n

If \u03b8 = 30\u00b0, verify that :<\/strong><\/p>\n\n\n\n

cos3\u03b8 = 4cos3<\/sup>\u03b8 \u2013 3cos\u03b8<\/strong>
Sol :
Given: \u03b8 =30o<\/sup>
Now, LHS = cos 3(30o<\/sup>)
\u21d2 cos 90o<\/sup>
= 0
Now, RHS = 4 cos3<\/sup> \u03b8 – 3 cos \u03b8
\u21d2 4 cos3<\/sup> (30o<\/sup>) – 3 cos (30o<\/sup>)
$\\Rightarrow 4\\left(\\frac{\\sqrt{3}}{2}\\right)^{3}-3\\left(\\frac{\\sqrt{3}}{2}\\right)$
$\\Rightarrow\\left(\\frac{3 \\sqrt{3}}{2}\\right)-\\left(\\frac{3 \\sqrt{3}}{2}\\right)$
= 0
\u2234 LHS = RHS
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 9 <\/h4>\n\n\n\n

If sin (A + B) = 1 and cos (A \u2013 B) = <\/strong>$\\frac{\\sqrt{3}}{2}$<\/strong>, then find A and B.<\/strong><\/p>\n\n\n\n

Sol :
Given : sin (A+B) =1
\u21d2 Sin(A+B) = sin (90 o<\/sup>) [\u2235 sin (90 o<\/sup>)=1]
On equating both the sides, we get
A + B = 90 o<\/sup> \u2026(1)
And $\\cos (A-B)=\\frac{\\sqrt{3}}{2}$
\u21d2 cos(A \u2013 B) = cos (30 o<\/sup>) $\\left[\\because \\cos \\left(30^{\\circ}\\right)=\\frac{\\sqrt{3}}{2}\\right]$<\/p>\n\n\n\n

On equating both the sides, we get
A \u2013 B = 30 o<\/sup> \u2026(2)<\/p>\n\n\n\n

On Adding Eq. (1) and (2), we get
2A = 120 o<\/sup>
\u21d2 A = 60 o<\/sup><\/p>\n\n\n\n

Now, Putting the value of A in Eq.(1), we get
60 o<\/sup> + B =90 o<\/sup>
\u21d2 B = 30 o<\/sup><\/p>\n\n\n\n

Hence, A = 60 o<\/sup> and B = 30 o<\/sup><\/p>\n\n\n\n

\n\n\n\n

Question 10 <\/h4>\n\n\n\n

If sin (A + B) = 1 and cos (A \u2013 B) = 1, find A and B.<\/strong><\/p>\n\n\n\n

Sol :
Given : sin (A+B) =1
\u21d2 Sin(A+B) = sin (90 o<\/sup>) [\u2235 sin (90 o<\/sup>) =1]
On equating both the sides, we get
A + B = 90 o<\/sup> \u2026(1)
And cos (A \u2013 B) = 1
\u21d2 cos(A \u2013 B) = cos (0 o<\/sup>) [\u2235 cos(0 o<\/sup>) = 1]
On equating both the sides, we get
A \u2013 B = 0 o<\/sup> \u2026(2)
On Adding Eq. (1) and (2), we get
2A = 90 o<\/sup>
\u21d2 A = 45 o<\/sup>
Now, Putting the value of A in Eq.(1), we get
45 o<\/sup> + B =90 o<\/sup>
\u21d2 B = 45 o<\/sup>
Hence, A = 45 o<\/sup> and B = 45 o<\/sup><\/p>\n\n\n\n

\n\n\n\n

Question 11 <\/h4>\n\n\n\n

If sin (A + B) = cos (A \u2013 B) = <\/strong>$\\frac{\\sqrt{3}}{2}$<\/strong>, fins A and B.<\/strong><\/p>\n\n\n\n

Sol :
Given : $\\sin (A+B)=\\frac{\\sqrt{3}}{2}$
\u21d2 Sin(A+B) = sin (60 o<\/sup>) $\\left[\\because \\sin (60 ^{\\circ})=\\frac{\\sqrt{3}}{2}\\right]$<\/p>\n\n\n\n

On equating both the sides, we get
A + B = 60 o<\/sup> \u2026(1)
And $\\cos (A-B)=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

\u21d2 cos(A \u2013 B) = cos (30 o<\/sup>) $\\left[\\because \\cos (30^{\\circ})=\\frac{\\sqrt{3}}{2}\\right]$<\/p>\n\n\n\n

On equating both the sides, we get
A \u2013 B = 30 o<\/sup> \u2026(2)<\/p>\n\n\n\n

On Adding Eq. (1) and (2), we get
2A = 90 o<\/sup>
\u21d2 A = 45 o<\/sup><\/p>\n\n\n\n

Now, Putting the value of A in Eq.(1), we get
45 o<\/sup> + B =60 o<\/sup>
\u21d2 B = 15 o<\/sup>
Hence, A = 45 o<\/sup> and B = 15 o<\/sup><\/p>\n\n\n\n

\n\n\n\n

Question 12 <\/h4>\n\n\n\n

If sin (A \u2013 B) = 1\/2, cos(A + B) = 1\/2; 0o<\/sup><A+B<90o<\/sup>; A > B, find A and B.<\/strong><\/p>\n\n\n\n

Sol :
Given : $\\sin (A-B)=\\frac{1}{2}$
\u21d2 Sin(A-B) = sin (30 o<\/sup>) $\\left[\\because \\sin (30^{\\circ})=\\frac{1}{2}\\right]$<\/p>\n\n\n\n

On equating both the sides, we get
A – B = 30 o<\/sup> \u2026(1)
And $\\cos (A+B)=\\frac{1}{2}$<\/p>\n\n\n\n

\u21d2 cos(A + B) = cos (60 o<\/sup>) $\\left[\\because \\cos (60^{\\circ})=\\frac{1}{2}\\right]$<\/p>\n\n\n\n

On equating both the sides, we get
A + B = 60 o<\/sup> \u2026(2)<\/p>\n\n\n\n

On Adding Eq. (1) and (2), we get
2A = 90 o<\/sup>
\u21d2 A = 45 o<\/sup><\/p>\n\n\n\n

Now, Putting the value of A in Eq.(2), we get
45 o<\/sup> + B =60 o<\/sup>
\u21d2 B = 15 o<\/sup>
Hence, A = 45 o<\/sup> and B = 15 o<\/sup><\/p>\n\n\n\n

\n\n\n\n

Question 13 A <\/h4>\n\n\n\n

Show by an example that<\/strong><\/p>\n\n\n\n

cos A \u2013 cos B \u2260 cos (A \u2013 B)<\/strong>
Sol :<\/p>\n\n\n\n

Let A = 60o<\/sup> and B = 30o<\/sup>, then
L.H.S. =cos A-cosB<\/p>\n\n\n\n

=cos60\u00b0-cos30\u00b0<\/p>\n\n\n\n

$=\\frac{1}{2}-\\frac{\\sqrt{3}}{2}=\\frac{1-\\sqrt{3}}{2}$<\/p>\n\n\n\n

R. H. S=cos(A-B)<\/p>\n\n\n\n

=cos(60\u00b0-30\u00b0)<\/p>\n\n\n\n

$=\\cos 30^{\\circ}=\\frac{\\sqrt{3}}{2}$<\/p>\n\n\n\n

\u2234 L.H.S. R.H.S<\/p>\n\n\n\n

\n\n\n\n

Question 13 B <\/h4>\n\n\n\n

Show by an example that<\/strong><\/p>\n\n\n\n

cos C + cos D \u2260 cos (C + D)<\/strong>
Sol :
Let C = 60o<\/sup> and D = 30o<\/sup>, then
L.H.S. = cos C + cos D = cos 60o<\/sup> + cos 30o<\/sup>
$=\\frac{1}{2}+\\frac{\\sqrt{3}}{2}=\\frac{1+\\sqrt{3}}{2}$
R. H. S. = cos (C+D) = cos (60o<\/sup> + 30o<\/sup>) = cos 90o<\/sup>= 0
\u2234 L.H.S. R.H.S<\/p>\n\n\n\n

\n\n\n\n

Question 13 C <\/h4>\n\n\n\n

Show by an example that<\/strong><\/p>\n\n\n\n

sin A + sin B \u2260 sin (A + B)<\/strong>
Sol :
Let A = 60o<\/sup> and B = 30o<\/sup>, then
L.H.S. = sin A + sin B = sin 60o<\/sup> + sin 30o<\/sup>
$=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}=\\frac{\\sqrt{3}+1}{2}$<\/p>\n\n\n\n

R. H. S. = sin (A + B) = sin (60o<\/sup> + 30o<\/sup>) = sin 90o<\/sup> =1
\u2234 L.H.S. R.H.S<\/p>\n\n\n\n

\n\n\n\n

Question 13 D <\/h4>\n\n\n\n

Show by an example that<\/strong><\/p>\n\n\n\n

sin A \u2013 sin B \u2260 sin (A \u2013 B)<\/strong>
Sol :
Let A = 60o<\/sup> and B = 30o<\/sup>, then
L.H.S. = sin A – sin B = sin 60o<\/sup> – sin 30o<\/sup>
$=\\frac{\\sqrt{3}}{2}-\\frac{1}{2}=\\frac{\\sqrt{3}-1}{2}$<\/p>\n\n\n\n

R. H. S. = sin (A – B) = sin (60o<\/sup> – 30o<\/sup>) = sin 30o<\/sup>
$=\\frac{1}{2}$
\u2234 L.H.S. \u2260R.H.S<\/p>\n\n\n\n

\n\n\n\n

Question 14 <\/h4>\n\n\n\n

In a right \u0394ABC hypotenuse AC = 10 cm and \u2220A = 60\u00b0, then find the length of the remaining sides.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Given: \u2220A = 60o<\/sup> and AC = 10cm
Now, $\\sin 60^{\\circ}=\\frac{\\text { Perpendicular }}{\\text { Hypotenuse }}=\\frac{\\mathrm{BC}}{\\mathrm{AC}}=\\frac{\\mathrm{BC}}{10}$<\/p>\n\n\n\n

Now, we know that $\\sin 60^{\\circ}=\\frac{\\sqrt{3}}{2}$
$\\Rightarrow \\frac{\\sqrt{3}}{2}=\\frac{B C}{10}$
\u21d2 BC = 5\u221a3 cm
In right angled \u2206ABC , we have
\u21d2 (AB)2<\/sup> + (BC)2<\/sup> =(AC)2<\/sup> [by using Pythagoras theorem]
\u21d2 (AB)2<\/sup> + (5\u221a3)2<\/sup> = (10)2<\/sup>
\u21d2 (AB)2<\/sup> +(25\u00d73) =100
\u21d2 (AB)2<\/sup> +75 = 100
\u21d2 (AB)2<\/sup> = 100 \u2013 75
\u21d2 (AB)2<\/sup> = 25
\u21d2 AB =\u221a25
\u21d2 AB = \u00b15
\u21d2 AB = 5cm [taking positive square root since, side cannot be negative]
\u2234 Length of the side AB = 5cm and BC =5\u221a3 cm<\/p>\n\n\n\n

\n\n\n\n

Question 15 <\/h4>\n\n\n\n

In a rectangle ABCD, BD : BC = 2 : \u221a3, then find \u2220BDC in degrees.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Given BD: BC = 2 : \u221a3
We have to find the \u2220BDC
We know that,
$\\sin \\theta=\\frac{\\text { Perpendicular }}{\\text { Hypotenuse }}$
$\\Rightarrow \\sin \\theta=\\frac{k \\sqrt{3}}{2 k}$
$\\Rightarrow \\sin \\theta=\\frac{\\sqrt{3}}{2}$
\u21d2 sin \u03b8 = sin 60o<\/sup>
\u21d2 \u03b8 = 60o<\/sup><\/p>\n\n\n\n

\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  Find the value of the following : (i) sin 30o + cos 60o(ii) sin2 45o+cos245o(iii) sin 30o + cos 60o \u2013 tan45o(iv) $\\sqrt{1+\\tan ^{2} 60^{\\circ}}$(v) tan 60o x cos30oSol :(i) sin 30o + cos 60oWe know that, $\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}>\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$ So,sin(30o) + cos(60o)$=\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{2}\\right)$=1 (ii) sin2 45o+cos245oWe know that,$\\sin \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$$\\cos \\left(45^{\\circ}\\right)=\\frac{1}{\\sqrt{2}}$ So, sin2 45o+cos245o$=\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}+\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}$$=\\left(\\frac{1}{2}\\right)+\\left(\\frac{1}{2}\\right)$=1 (iii) sin 30o + cos 60o \u2013 tan45o$\\sin \\left(30^{\\circ}\\right)=\\frac{1}{2}$$\\cos \\left(60^{\\circ}\\right)=\\frac{1}{2}$tan(45o)=1 So,sin […]<\/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-624094","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 4.2 - 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