{"id":625827,"date":"2023-09-08T06:38:28","date_gmt":"2023-09-08T06:38:28","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=625827"},"modified":"2023-09-08T06:38:34","modified_gmt":"2023-09-08T06:38:34","slug":"kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/","title":{"rendered":"KC Sinha: Exercise 6.2 – Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928"},"content":{"rendered":"\n\n\n\n\n

(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0
Sol :
L.H.S
=sin65\u00b0+cos65\u00b0
=sin65\u00b0+sin(90\u00b0-65\u00b0)
=sin65\u00b0+sin25\u00b0<\/p>\n\n\n\n

\u2235$\\left[\\begin{array}{c} \\sin C+\\sin D =2 \\sin \\frac{C+D}{2} \\cos \\frac{C-D}{2}\\end{array}\\right]$<\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) $\\frac{\\cos 10^{\\circ}-\\sin 10^{\\circ}}{\\cos 10^{\\circ}+\\sin 10^{\\circ}}$
Sol :
L.H.S
=$\\frac{\\cos 10^{\\circ}-\\sin 10^{\\circ}}{\\cos 10^{\\circ}+\\sin 10^{\\circ}}$<\/p>\n\n\n\n

$=\\frac{\\sin (90^{\\circ}-10^{\\circ})-\\sin 10^{\\circ}}{\\sin (90^{\\circ}-10^{\\circ})+\\sin 10^{\\circ}}$<\/p>\n\n\n\n

\u2235$\\left[\\begin{array}{c} \\sin C-\\sin D =2 cos \\frac{C+D}{2} sin \\frac{C-D}{2}\\end{array}\\right]$<\/p>\n\n\n\n

\u2235$\\left[\\begin{array}{c} \\sin C+\\sin D =2 \\sin \\frac{C+D}{2} \\cos \\frac{C-D}{2}\\end{array}\\right]$<\/p>\n\n\n\n

 $=\\frac{2 \\cos \\frac{80^{\\circ}+10^{\\circ}}{2} \\sin \\frac{80^{\\circ}-10^{\\circ}}{2}}{2 \\sin \\frac{80^{\\circ}+10^{\\circ}}{2} \\cos \\frac{80^{\\circ}-10^{\\circ}}{2}}$<\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) cos80\u00b0+cos40\u00b0-cos20\u00b0=0
Sol :
L.H.S
=cos80\u00b0+cos40\u00b0-cos20\u00b0<\/p>\n\n\n\n

$=2 Cos\\frac{ 80^{\\circ}+40^{\\circ}}{2} \\cos \\frac{80^{\\circ}-40^{\\circ}}{2}-\\cos 20^{\\circ}$<\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin10\u00b0+sin20\u00b0+sin40\u00b0+sin50\u00b0=sin70\u00b0+sin80\u00b0
Sol :
L.H.S
=sin10\u00b0+sin20\u00b0+sin40\u00b0+sin50\u00b0<\/p>\n\n\n\n

 $=2 \\sin \\frac{50^{\\circ}+40^{\\circ}}{2} \\cdot \\cos \\frac{50^{\\circ}-10^{\\circ}}{2}+2 \\sin \\frac{40^{\\circ}+20^{\\circ}}{2} cos \\frac{40^{\\circ}-20^{\\circ}}{2}$<\/p>\n\n\n\n

$=2 \\times \\dfrac{1}{2} \\times \\cos 20^{\\circ}+2 \\times \\dfrac{1}{2} \\cos 10^{\\circ}$<\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) $\\cos \\frac{\\pi}{5}+\\cos \\frac{2 \\pi}{5}+\\cos \\frac{6 \\pi}{5}+\\cos \\frac{7 \\pi}{5}=0$<\/p>\n\n\n\n

$=\\cos \\frac{\\pi}{5}+\\cos \\frac{2 \\pi}{5}+\\cos \\left(\\pi+\\frac{\\pi}{5}\\right)+\\cos \\left(\\pi+\\frac{2 \\pi}{5}\\right)$<\/p>\n\n\n\n

$=\\cos \\frac{\\pi}{5}+\\cos \\frac{2 \\pi}{5}-\\cos \\frac{\\pi}{5}-\\cos \\frac{2 \\pi}{5}$<\/p>\n\n\n\n

\u2235$\\left[\\begin{array}{c} \\sin C-\\sin D =2 cos \\frac{C+D}{2} sin \\frac{C-D}{2}\\end{array}\\right]$<\/p>\n\n\n\n

\u2235$\\left[\\begin{array}{c} \\sin C+\\sin D =2 \\sin \\frac{C+D}{2} \\cos \\frac{C-D}{2}\\end{array}\\right]$<\/p>\n\n\n\n

$=\\frac{2 cos \\frac{7 x+5x}{2} \\cos \\frac{7 x-5 x}{2}}{2 \\cos \\frac{7 x+5}{2} \\sin \\frac{7 x-5}{2}}$<\/p>\n\n\n\n

$=\\frac{\\cos \\frac{2 x}{2}}{\\tan \\frac{2 x}{2}}$<\/p>\n\n\n\n

$=\\frac{\\cos x}{\\sin x}=\\cot x$<\/p>\n\n\n\n

(iv)<\/strong> cot 4x(six5x+sin3x)=cotx(sin5x-sin3x)
Sol :
L.H.S
=cot 4x(six5x+sin3x)<\/p>\n\n\n\n

$=\\cot 4 x \\cdot 2 \\sin \\frac{5 x+3 x}{2} \\cos \\frac{5 x-3 x}{2}$<\/p>\n\n\n\n

$=\\frac{\\cos x}{\\sin 4 x} \\times 2 \\sin 4 x \\times \\cos x$<\/p>\n\n\n\n

=2cos4x.cosx<\/p>\n\n\n\n

R.H.S
=cotx(sin5x-sin3x)<\/p>\n\n\n\n

$=\\cot x \\cdot 2 \\cos \\frac{5 x+3 x}{2} \\sin \\frac{5 x-3 x}{2}$<\/p>\n\n\n\n

$=\\dfrac{\\cos x}{\\sin x} \\times 2 \\cos 4 x \\sin x$<\/p>\n\n\n\n

=2cos4x.cosx<\/p>\n\n\n\n

(v)<\/strong> $\\frac{\\sin x-\\sin 3 x}{\\sin ^{2} x-\\cos ^{2} x}=2 \\sin x$
Sol :
L.H.S
$\\frac{\\sin x-\\sin 3 x}{\\sin ^{2} x-\\cos ^{2} x}$<\/p>\n\n\n\n

$=\\frac{-(\\sin 3 x-\\sin x)}{-\\left(\\cos ^{2} x-\\sin ^{2} x\\right)}$<\/p>\n\n\n\n

\u2235 cos2<\/sup>A-sin2<\/sup>B=cos(A+B).cos(A-B)<\/p>\n\n\n\n

\u2235 cos2<\/sup>x-sin2<\/sup>x=cos2x<\/p>\n\n\n\n

$=\\frac{2 \\cos \\frac{3 x+x}{2} \\sin \\frac{3 x-x}{2}}{\\cos (x+x) \\cos (x-x)}$<\/p>\n\n\n\n

$=\\frac{2 \\cos 2 x \\sin x}{\\cos 2 x \\cos 0}$<\/p>\n\n\n\n

$=\\dfrac{2 \\sin x}{1}$<\/p>\n\n\n\n

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

Question 7<\/strong><\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that)<\/p>\n\n\n\n

(i)<\/strong> $\\frac{\\sin x-\\sin y}{\\cos x+\\cos y}=\\tan \\frac{x-y}{2}$<\/p>\n\n\n\n

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

L.H.S<\/p>\n\n\n\n

$\\frac{\\sin x-\\sin y}{\\cos x+\\cos y}$<\/p>\n\n\n\n

$=\\frac{2 \\cos \\frac{x+y}{2} \\sin \\frac{x-y}{2}}{2 \\cos \\frac{x+y}{2} \\cos \\frac{x-y}{2}}$<\/p>\n\n\n\n

$=\\tan \\dfrac{x – y}{2}$<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{\\sin x+\\sin y}{\\sin x-\\sin y}=\\tan \\left(\\frac{x+y}{2}\\right) \\cdot \\cot \\left(\\frac{x-y}{2}\\right)$<\/p>\n\n\n\n

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

Question 8<\/strong><\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) <\/p>\n\n\n\n

(i) sin2x+2sin4x+sin6x=4cos2<\/sup>x six4x<\/p>\n\n\n\n

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

=sin2x+2sin4x+sin6x<\/p>\n\n\n\n

=2sin4x+(sin6x+sin2x)<\/p>\n\n\n\n

=$2 \\sin 4 x+2 \\sin \\frac{6 x+2 x}{2} \\cos \\frac{6 x-2 x}{2}$<\/p>\n\n\n\n

=2sin4x+2sin4x.cos2x<\/p>\n\n\n\n

=2sin4x(1+cos2x)<\/p>\n\n\n\n

[\u2235 1+cos2x=2cos^2x]<\/p>\n\n\n\n

=4cos2<\/sup>x.sin4x<\/p>\n\n\n\n

(ii) sin x+sin 3x+sin 5x+sin 7x=4 cos x cos 2x sin 4 x<\/p>\n\n\n\n

Sol :
L.H.S
= sin x+sin 3x+sin 5x+sin 7x<\/p>\n\n\n\n

=(sin7x+sinx)+(sin5x+sin3x)<\/p>\n\n\n\n

=$=2 \\sin \\frac{7 x+x}{2} \\cos \\frac{7 x-x}{2}+2 \\sin \\frac{5 x+3 x}{2} \\cos \\frac{5 x-3 x}{2}$<\/p>\n\n\n\n

=2sin4x.cox3x+2sin4x.cosx<\/p>\n\n\n\n

=2sin4x(cos3x+cosx)<\/p>\n\n\n\n

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

=4sin4x.cos2x.cosx<\/p>\n\n\n\n

=4cosx.cos2x.sin4x<\/p>\n\n\n\n

Question 9<\/strong><\/p>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) <\/p>\n\n\n\n

(i)<\/strong> $\\frac{\\cos 4 \\theta+\\cos 3 \\theta+\\cos 2 \\theta}{\\sin 4 \\theta+\\sin 3 \\theta+\\sin 2 \\theta}=\\cot 3 \\theta$<\/p>\n\n\n\n

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

L.H.S<\/p>\n\n\n\n

=$\\dfrac{\\cos 4 \\theta+\\cos 3 \\theta+\\cos 2 \\theta}{\\sin 4 \\theta+\\sin 3 \\theta+\\sin 2 \\theta}$<\/p>\n\n\n\n

$=\\dfrac{(\\cos 4 \\theta+\\cos 2 \\theta)+\\cos 3 \\theta}{(\\sin 4 \\theta+\\sin 2 \\theta)+\\sin 3 \\theta}$<\/p>\n\n\n\n

$=\\dfrac{2 \\cos \\frac{4 \\theta+2 \\theta}{2} \\cos \\frac{4 \\theta-2 \\theta}{2}+\\cos 3 \\theta}{2 \\sin \\frac{4 \\theta+2 \\theta}{2} \\cos \\frac{4 \\theta-2 \\theta}{2}+\\sin 3 \\theta}$<\/p>\n\n\n\n

$=\\dfrac{2 \\cos 3 \\theta \\cos \\theta+\\cos 3 \\theta}{2 \\sin 3 \\theta \\cos \\theta+\\sin 3 \\theta}$<\/p>\n\n\n\n

$=\\frac{\\cos 3 \\theta\\left(2 \\cos \\theta+1\\right)}{\\sin 3 \\theta(2 \\cos \\theta+1)}$<\/p>\n\n\n\n

=cot3\u03b8<\/p>\n\n\n\n

(ii)<\/strong> $\\dfrac{\\sin 5 \\theta-2 \\sin 3 \\theta+\\sin \\theta}{\\cos 5 \\theta-\\cos \\theta}=\\tan \\theta$<\/p>\n\n\n\n

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

L.H.S<\/p>\n\n\n\n

=$\\dfrac{\\sin 5\\theta-2 \\sin 3 \\theta+\\sin \\theta}{\\cos 5 \\theta-\\cos \\theta}$<\/p>\n\n\n\n

$=\\frac{(\\sin 5 \\theta+\\sin \\theta)-2 \\sin 3 \\theta}{\\cos 5 \\theta-\\cos \\theta}$<\/p>\n\n\n\n

\u2235$\\left[\\begin{array}{c} \\cos C-\\cos D =-2sin \\frac{C+D}{2} sin \\frac{C-D}{2}\\end{array}\\right]$<\/p>\n\n\n\n

$=\\frac{2 \\tan \\frac{5 \\theta+\\theta}{2} \\cos \\frac{5 \\theta-\\theta}{2}-2 \\sin 3 \\theta}{-2 \\sin \\frac{5 \\theta+\\theta}{2} \\sin \\frac{5 \\theta-\\theta}{2}}$<\/p>\n\n\n\n

$=\\frac{2 \\sin 3 \\theta \\cos 2 \\theta-2 \\sin 3 \\theta}{-2 \\sin 3 \\theta \\sin 2 \\theta}$<\/p>\n\n\n\n

$=\\frac{2 sin 30(\\cos 2 \\theta-1)}{-2 \\sin 3 \\theta.sin 2\\theta}$<\/p>\n\n\n\n

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

[\u2235 1-cos2x=2sin2<\/sup>x<\/p>\n\n\n\n

sin 2x=2sinx cosx]<\/p>\n\n\n\n

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

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

=tan\u03b8<\/p>\n\n\n\n

(iii)<\/strong> $\\frac{(\\sin 7 x+\\sin 5 x)+(\\sin 9 x+\\sin 3 x)}{(\\cos 7 x+\\cos 5 x)+(\\cos 9 x+\\cos 3 x)}=\\tan 6 x$<\/p>\n\n\n\n

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

L.H.S<\/p>\n\n\n\n

=$\\frac{(\\sin 7 x+\\sin 5 x)+(\\sin 9 x+\\sin 3 x)}{(\\cos 7 x+\\cos 5 x)+(\\cos 9 x+\\cos 3 x)}$<\/p>\n\n\n\n

$=\\frac{2 \\sin \\frac{7 x+5 x}{2} \\cos 7 \\frac{x-5 x}{2}+2 \\sin \\frac{9 x+3 x}{2} \\cos \\frac{9 x-3 x}{2}}{2 \\cos \\frac{7 x+5 x}{2} \\cos \\frac{7 x-5 x}{2}+2 \\cos \\frac{9 x+3 x}{2} \\cos \\frac{9 x-3 x}{2}}$<\/p>\n\n\n\n

$=\\frac{2 \\sin 6 x \\cos x+2 \\sin 6 x \\cos 3 x}{2 \\cos 6 x \\cos x+2 \\cos 6 x \\cos 3 x}$<\/p>\n\n\n\n

$=\\frac{2 \\sin 6 x\\left(\\cos x+\\cos 3 x\\right)}{2 \\cos 6 x(\\cos x+\\cos 3 x)}$<\/p>\n\n\n\n

=tan 6x<\/p>\n\n\n\n

(i)
Sol :
$\\sin \\alpha-\\sin \\beta=\\frac{1}{3}$<\/p>\n\n\n\n

$\\because \\sin C-\\sin B=2 \\cos \\frac{C+D}{2} \\sin \\frac{C-D}{2}$<\/p>\n\n\n\n

$2 \\cos \\frac{\\alpha+\\beta}{2} \\sin \\frac{\\alpha-\\beta}{2}=\\frac{1}{3}$..(i)<\/p>\n\n\n\n

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

$\\because \\cos C-\\cos D=2 \\sin \\frac{C+D}{2} \\sin \\frac{D-C}{2}$<\/p>\n\n\n\n

$2 \\sin \\frac{\\beta+\\alpha}{2} \\sin\\frac{ \\alpha-\\beta}{2}=\\frac{1}{2}$..(ii)<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

$\\frac{2 \\cos \\frac{\\alpha+\\beta}{2} \\sin \\frac{\\alpha-\\beta}{2}}{2 \\sin \\frac{\\beta+\\gamma}{2} \\sin \\frac{\\alpha-\\beta}{2}}=\\frac{\\frac{1}{3}}{\\frac{1}{2}}$<\/p>\n\n\n\n

$\\cot\\frac{\\alpha+\\beta}{2}=\\frac{2}{3}$<\/p>\n\n\n\n

(ii)<\/strong>
Sol :<\/p>\n\n\n\n

$\\operatorname{cosec} A+\\sec A=\\operatorname{cosec} B+\\sec B$<\/p>\n\n\n\n

secA-secB=cosecB-cosecA<\/p>\n\n\n\n

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

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

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

$\\tan A \\tan B=\\frac{2 \\cos \\frac{A+B}{2}\\sin\\frac{A-B}{2}}{2 \\sin \\frac{B+A}{2} \\sin \\frac{A-B}{2}}$<\/p>\n\n\n\n

$\\tan A \\tan B=\\cot \\frac{A+B}{2}$<\/p>\n\n\n\n

Question 11<\/strong><\/p>\n\n\n\n

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

$\\sec (\\theta+\\alpha)+\\sec (\\theta-\\alpha)=2 \\sec \\theta$<\/p>\n\n\n\n

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

$\\frac{\\cos (\\theta-\\alpha)+\\cos (\\theta+\\alpha)}{\\cos (\\theta+\\alpha) \\cos (\\theta-\\alpha)}=\\frac{2}{\\cos \\theta}$<\/p>\n\n\n\n

\u2235 cos(A+B)+cos(A-B)=2cosA.cosB<\/p>\n\n\n\n

\u2235 cos(A+B).cos(A-B)=cos2<\/sup>A-sin2<\/sup>A<\/p>\n\n\n\n

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

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

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

$1^{2}-\\cos ^{2} \\alpha=\\cos ^{2} \\theta(1-\\cos \\alpha)$<\/p>\n\n\n\n

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

$1+\\cos \\alpha=\\cos ^{2} \\theta$<\/p>\n\n\n\n

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

$\\sin 25^{\\circ} \\cos 115^{\\circ}=\\frac{1}{2}\\left(\\sin 140^{\\circ}-1\\right)$
Sol :
L.H.S
$\\sin 25^{\\circ} \\cos 115^{\\circ}$<\/p>\n\n\n\n

[\u22352sinAcosB=sin(A+B)+sin(A-B)]<\/p>\n\n\n\n

$=\\frac{1}{2}(2 \\sin25^{\\circ} \\cos 115^{\\circ})$<\/p>\n\n\n\n

$=\\frac{1}{2}[\\sin (25^{\\circ}+115^{\\circ})+\\sin (25^{\\circ}-115^{\\circ})]$<\/p>\n\n\n\n

$=\\frac{1}{2}[\\sin 140^{\\circ}+\\sin (-90^{\\circ})]$<\/p>\n\n\n\n

$=\\frac{1}{2}[\\sin 140^{\\circ}-\\sin 90^{\\circ}]$<\/p>\n\n\n\n

$=\\frac{1}{2}[\\sin 140^{\\circ}-1]$<\/p>\n\n\n\n

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

[]
$\\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 60^{\\circ} \\sin 80^{\\circ}=\\frac{3}{16}$
Sol :
L.H.S
$\\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 60^{\\circ} \\sin 80^{\\circ}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{2} \\sin 20^{\\circ} \\times \\frac{1}{2}(2 \\sin 80^{\\circ} \\sin 40^{\\circ})$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4} \\sin 20^{\\circ}\\left(\\cos \\left(80^{\\circ}-40\\right)-\\cos \\left(80^{\\circ}+40^{\\circ}\\right)\\right]$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4} \\sin 20^{\\circ}\\left[\\cos 40^{\\circ}-\\cos 120^{\\circ}\\right]$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4} \\sin 2 0^{\\circ}\\left[\\cos 40^{\\circ}-\\cos (180^{\\circ}-10^{\\circ})\\right]$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4} \\sin 20^{\\circ}\\left[\\cos 40^{\\circ}-\\left(-\\cos 60^{\\circ}\\right)\\right]$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4} \\sin 20^{\\circ}\\left[\\cos 40^{\\circ}+\\frac{1}{2}\\right]$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{4} \\sin 20^{\\circ} \\cos 40^{\\circ}+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{8}(2 \\sin 20^{\\circ} \\cos 40^{\\circ})+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{8}[\\sin (20^{\\circ}+40^{\\circ})+\\sin (20^{\\circ}-40^{\\circ})]+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{8}\\left(\\sin 60^{\\circ}+\\sin \\left(-20^{\\circ}\\right)\\right]+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{3}}{8}\\left[\\frac{\\sqrt{3}}{2}-\\sin 20^{\\circ}\\right]+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{3}{16}-\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}+\\frac{\\sqrt{3}}{8} \\sin 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{3}{16}$<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

[]
$\\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 60^{\\circ} \\cos 80^{\\circ}=\\frac{1}{16^{\\circ}}$
Sol :
L.H.S<\/p>\n\n\n\n

$\\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 60^{\\circ} \\cos 80^{\\circ}$<\/p>\n\n\n\n

$=\\frac{1}{2} \\cos 20^{\\circ} \\times \\frac{1}{2}(2 \\cos 80^{\\circ} \\cos 40^{\\circ})$<\/p>\n\n\n\n

[\u2235 2cosAcosB=cos(A+B)+cos(A-B)]<\/p>\n\n\n\n

$=\\frac{1}{4} \\cos 20^{\\circ}[\\cos (80^{\\circ}+40^{\\circ})+\\cos (80^{\\circ}-40^{\\circ})]$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cos 20^{\\circ}\\left[\\cos 120^{\\circ}+\\cos 40^{\\circ}\\right]$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cos 20^{\\circ}\\left[\\cos (180^{\\circ}-60^{\\circ})+\\cos 40^{\\circ}\\right]$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cos 20^{\\circ}[-\\cos 60^{\\circ}+\\cos 40^{\\circ}]$<\/p>\n\n\n\n

$=\\frac{1}{4} \\cos 20^{\\circ}\\left[-\\frac{1}{2}+\\cos 40^{\\circ}\\right]$<\/p>\n\n\n\n

$=-\\frac{1}{8} \\cos 20^{\\circ}+\\frac{1}{4} \\cos 40^{\\circ} \\cos 20^{\\circ}$<\/p>\n\n\n\n

$=-\\frac{1}{8} \\cos 20^{\\circ}+\\frac{1}{8}(2 \\cos 40^{\\circ} \\cos 20^{\\circ})$<\/p>\n\n\n\n

$=-\\frac{1}{8} \\cos 20^{\\circ}+\\frac{1}{8}[\\cos (40^{\\circ}+20^{\\circ})+\\cos (40^{\\circ}-20^{\\circ})]$<\/p>\n\n\n\n

$=-\\frac{1}{8} \\cos 20^{\\circ}+\\frac{1}{8}[\\cos 60^{\\circ}+\\cos 20^{\\circ}]$<\/p>\n\n\n\n

$=-\\frac{1}{8} \\cos 20^{\\circ}+\\frac{1}{8} \\cos 60^{\\circ}+\\frac{1}{8} \\cos 20^{\\circ}$<\/p>\n\n\n\n

$=\\frac{1}{8} \\times \\frac{1}{2}$<\/p>\n\n\n\n

$=\\frac{1}{16}$<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

Question 17<\/h4>\n\n\n\n

$4 \\cos \\theta \\cos \\left(\\frac{\\pi}{3}+\\theta\\right) \\cos \\left(\\frac{\\pi}{3}-\\theta\\right)=\\cos 3 \\theta$
Sol :
L.H.S
$4 \\cos \\theta \\cos \\left(\\frac{\\pi}{3}+\\theta\\right) \\cos \\left(\\frac{\\pi}{3}-\\theta\\right)$<\/p>\n\n\n\n

$=2 \\cos \\theta \\cdot 2 \\cos \\left(\\frac{\\pi}{3}+\\theta\\right) \\cos \\left(\\frac{\\pi}{3}-\\theta\\right)$<\/p>\n\n\n\n

[\u2235 2cosA.cosB=cos(A+B)+cos(A-B)]<\/p>\n\n\n\n

$=2 \\cos \\theta\\left[\\cos \\left\\{\\left(\\frac{\\pi}{3}+\\theta\\right)+\\left(\\frac{\\pi}{3}-\\theta\\right)\\right\\}\\right.+\\cos \\left\\{\\left(\\frac{\\pi}{3}+\\theta\\right)-\\left(\\frac{\\pi}{2}-\\theta \\right)\\right]$<\/p>\n\n\n\n

$=2 \\cos \\theta\\left[\\cos \\left(\\frac{\\pi}{3}+\\theta+\\frac{\\pi}{3}-\\theta\\right)+\\cos \\left(\\frac{\\pi}{3}+\\theta-\\frac{\\pi}{3}+\\theta\\right)\\right.$<\/p>\n\n\n\n

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

$=2 \\cos \\theta\\left[\\cos \\left(\\pi-\\frac{\\pi}{3}\\right)+\\cos 2 \\theta\\right]$<\/p>\n\n\n\n

$=2 \\cos \\theta\\left[-\\cos \\frac{\\pi}{3}+\\cos 2 \\theta\\right]$<\/p>\n\n\n\n

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

$=-\\cos \\theta+2 \\cos 2 \\theta \\cos \\theta$<\/p>\n\n\n\n

$=-cos \\theta+\\cos (2 \\theta+\\theta)+\\cos (2 \\theta-\\theta)$<\/p>\n\n\n\n

$=-\\cos \\theta+\\cos 3 \\theta+\\cos \\theta$<\/p>\n\n\n\n

Question 18<\/h4>\n\n\n\n

$\\tan \\theta \\tan \\left(60^{\\circ}+\\theta\\right) \\tan \\left(60^{\\circ}-\\theta\\right)=\\tan 3 \\theta$
Sol :
L.H.S
$\\tan \\theta \\tan (60+\\theta) \\tan (60-\\theta)$<\/p>\n\n\n\n

$=\\frac{\\sin \\theta \\sin (60^{\\circ}+\\theta) \\sin (60^{\\circ}-\\theta)}{\\cos \\theta \\cos (60^{\\circ}+\\theta) \\cos (60^{\\circ}-\\theta)}$<\/p>\n\n\n\n

[\u2235 sin(A+B)sin(A-B=sin2<\/sup>A.sin2<\/sup>B]<\/p>\n\n\n\n

[\u2235 cos(A+B)cos(A-B)=cos2<\/sup>B-sin2<\/sup>A]<\/p>\n\n\n\n

$=\\frac{\\sin \\theta\\left[\\sin ^{2} 60^{\\circ}-\\sin ^{2} \\theta\\right]}{\\cos \\theta\\left[\\cos ^{2} \\theta-\\sin ^{2} 60^{\\circ}\\right]}$<\/p>\n\n\n\n

$=\\frac{\\sin \\theta\\left[\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\sin ^{2} \\theta\\right]}{\\cos \\theta\\left[\\cos ^{2} \\theta-\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}\\right]}$<\/p>\n\n\n\n

$=\\frac{\\sin \\theta\\left[\\frac{3}{4}-\\sin ^{2} \\theta\\right]}{\\cos \\theta\\left[\\cos ^{2} \\theta-\\frac{3}{4}\\right]}$<\/p>\n\n\n\n

$=\\frac{\\sin \\theta\\left[\\frac{3-4 \\sin ^{2} \\theta}{4}\\right]}{\\cos \\theta\\left[\\frac{4 \\cos ^{2} \\theta-3}{4}\\right]}$<\/p>\n\n\n\n

$=\\frac{3 \\sin \\theta-4 \\sin 3 \\theta}{4 \\cos ^{3} \\theta-3 \\cos \\theta}$<\/p>\n\n\n\n

$=\\frac{\\sin 3 \\theta}{\\cos 3 \\theta}=\\tan 3 \\theta$<\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

Sol :
$\\cos \\alpha \\cos \\beta=\\frac{1}{2} \\times 2 \\cos \\alpha \\cos \\beta$<\/p>\n\n\n\n

$\\cos \\alpha \\cos \\beta=\\frac{1}{2}[\\cos (\\alpha+\\beta)+\\cos (\\alpha-\\beta)]$<\/p>\n\n\n\n

$\\cos \\alpha \\cos \\beta=\\frac{1}{2}[\\cos 90+\\cos (\\alpha-\\beta)]$<\/p>\n\n\n\n

$\\cos \\alpha \\cos \\beta=\\frac{1}{2} \\cos (\\alpha-\\beta)$<\/p>\n\n\n\n

$(\\cos (\\alpha-\\beta) \\leq 1]$<\/p>\n\n\n\n

\u2234 $\\cos \\alpha \\cos \\beta \\leq \\frac{1}{2}(1)$<\/p>\n\n\n\n

$\\cos \\alpha \\cos \\beta \\leq \\frac{1}{2}$<\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

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

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

[]<\/p>\n\n\n\n

$\\sin \\alpha=\\pm \\sqrt{1-\\cos ^{2} \\alpha}$<\/p>\n\n\n\n

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

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

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

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

$\\sin \\alpha=+\\frac{1}{\\sqrt{2}}$ , $\\sin \\alpha=-\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

L.H.S<\/p>\n\n\n\n

$\\tan \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\alpha-\\beta}{2}$<\/p>\n\n\n\n

$=\\frac{2 \\sin \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\alpha-\\beta}{2}}{2 cos \\frac{\\alpha+\\beta}{2} \\sin \\frac{\\alpha-\\beta}{2}}$<\/p>\n\n\n\n

$=\\frac{\\sin \\alpha+\\sin \\beta}{\\sin \\alpha-\\sin \\beta}$<\/p>\n\n\n\n

CASE-I
$\\sin \\alpha=\\frac{1}{\\sqrt{2}}, \\quad \\sin \\beta=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

$\\Rightarrow \\frac{\\sin \\alpha+\\sin A}{\\sin \\alpha-\\sin \\beta}$<\/p>\n\n\n\n

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

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

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

[]<\/p>\n\n\n\n

$=\\frac{3+\\sqrt{6}+\\sqrt{6}+2}{(\\sqrt{3})^{2}-(\\sqrt{2})^{2}}$<\/p>\n\n\n\n

$=\\frac{5+2 \\sqrt{6}}{3-2}$<\/p>\n\n\n\n

$=5+2 \\sqrt{6}$<\/p>\n\n\n\n

CASE-II<\/p>\n\n\n\n

$\\sin \\alpha=-\\frac{1}{\\sqrt{2}} \\quad, \\sin \\beta=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

$\\frac{\\sin \\alpha+\\sin \\beta}{\\sin \\alpha-\\sin \\beta}=5-2 \\sqrt{6}$<\/p>\n\n\n\n

Question 21<\/h4>\n\n\n\n

Sol :
$x \\cos \\theta=y \\cos \\left(\\theta+\\frac{2 \\pi}{3}\\right)=z \\cos \\left(\\theta+\\frac{4 \\pi}{3}\\right)=k$<\/p>\n\n\n\n

$\\cos \\theta=\\frac{k}{x}$  , $\\cos \\left(\\theta+\\frac{2 \\pi}{3}\\right)=\\frac{k}{y}$ , $\\cos \\left(\\theta+\\frac{4 \\pi}{3}\\right)=\\frac{k}{z}$<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

$\\cos \\theta+\\cos \\left(\\theta+\\frac{2 \\pi}{3}\\right)+\\cos \\left(\\theta+\\frac{4 \\pi}{3}\\right)=\\frac{k}{x}+\\frac{k}{y}+\\frac{k}{z}$<\/p>\n\n\n\n

$\\cos \\theta+\\cos \\left[\\pi-\\left(\\frac{\\pi}{3}-\\theta\\right)\\right]+\\cos \\left[\\pi+\\left(\\frac{\\pi}{3}+\\theta\\right)\\right]=k\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)$<\/p>\n\n\n\n

$\\cos \\theta-\\cos \\left(\\frac{\\pi}{3}-\\theta\\right)-\\cos \\left(\\frac{\\pi}{3}+\\theta\\right)=k\\left(\\frac{y z+x z+x y}{x y z}\\right)$<\/p>\n\n\n\n

$\\cos \\theta-\\left[\\cos \\left(\\frac{\\pi}{3}-\\theta\\right)+\\cos \\left(\\frac{\\pi}{3}+\\theta\\right)\\right]=k \\frac{\\left(xy+yz+zx\\right)}{xyz}$<\/p>\n\n\n\n

$\\cos \\theta-\\left[2 \\cos \\frac{\\pi}{3} \\cos \\theta\\right]= \\frac{k(x y+y z+z x)}{x y z}$<\/p>\n\n\n\n

$\\cos \\theta-\\left[2 \\times \\frac{1}{2} \\cos \\theta\\right]=\\frac{k(x y+yz+z x)}{x y z}$<\/p>\n\n\n\n

$0=k\\left(\\frac{x y+y z+z x}{x y z}\\right)$<\/p>\n\n\n\n

0=xy+yz+zx<\/p>\n\n\n\n

Question 22<\/h4>\n\n\n\n

Sol :
$y \\sin \\phi=x \\sin (2 \\theta+\\phi)$<\/p>\n\n\n\n

$\\frac{y}{x}=\\frac{\\sin (2 \\theta+\\phi)}{\\sin \\phi}$<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

$\\frac{y+x}{y-x}=\\frac{\\sin (2 \\theta+\\phi)+\\sin \\phi}{\\sin (2 \\theta+\\phi)-\\sin \\phi}$<\/p>\n\n\n\n

$\\frac{x+y}{y-x}=\\frac{2 \\sin \\frac{2 \\theta+\\phi+\\phi}{2} \\cos \\frac{2 \\theta+\\phi-\\phi}{2}}{2 \\cos \\frac{2 \\theta+\\phi+\\phi}{2} \\sin \\frac{2 \\theta+\\phi-\\phi}{2}}$<\/p>\n\n\n\n

$\\frac{x+y}{y-x}=\\frac{\\sin \\frac{2(\\theta+\\phi)}{2}\\cos \\frac{2\\theta}{2}}{\\cos \\frac{2(\\theta+\\phi)}{2} \\sin \\frac{2 \\theta}{2}}$<\/p>\n\n\n\n

$\\frac{x+y}{y-x}=\\tan (\\theta+\\phi) \\cot \\theta$<\/p>\n\n\n\n

$\\frac{x+y}{\\tan (\\theta+\\phi)}=(y-x) \\cot \\theta$<\/p>\n\n\n\n

$(x+y) \\cot (\\theta+\\phi)=(y-x) \\cot \\theta$<\/p>\n\n\n\n

Question 23<\/h4>\n\n\n\n

Sol :
$\\cos (\\alpha+\\beta) \\sin (\\gamma+\\delta)=\\cos (\\alpha-\\beta) \\sin (\\gamma-\\delta)$<\/p>\n\n\n\n

$\\frac{\\cos (\\alpha+\\beta)}{\\cos (\\alpha-\\beta)}=\\frac{\\sin (\\gamma-\\delta)}{\\sin (\\gamma+\\delta)}$<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

$\\frac{\\cos (\\alpha+\\beta)+\\cos (\\alpha-\\beta)}{\\cos (\\alpha+\\beta)-\\cos (\\alpha-\\beta)}=\\frac{\\sin (\\gamma-\\delta)+\\sin (\\gamma+\\delta)}{\\sin (\\gamma-\\delta)-\\sin (\\gamma+\\delta)}$<\/p>\n\n\n\n

$\\frac{2 \\cos \\alpha \\cos \\beta}{-2 \\sin \\alpha \\sin \\beta}=\\frac{2 \\sin \\gamma \\cos \\delta}{-2 \\cos \\gamma \\sin \\delta}$<\/p>\n\n\n\n

$\\cot \\alpha \\cot \\beta=\\tan \\gamma \\cot \\delta$<\/p>\n\n\n\n

$\\frac{\\cot \\alpha \\cos \\beta}{\\tan \\gamma}=\\cos \\delta$<\/p>\n\n\n\n

$\\cot \\alpha \\cot \\beta \\cot \\gamma=\\cot\\delta$<\/p>\n\n\n\n

Question 24<\/h4>\n\n\n\n

Sol :
$\\frac{\\cos (A-B)}{\\cos (A+B)}+\\frac{\\cos (C+D)}{\\cos (C-D)}=0$<\/p>\n\n\n\n

$\\frac{\\cos (A-B)}{\\cos (A+B)}=\\frac{-\\cos (C+D)}{\\cos (C-D)}$<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

$\\frac{\\cos (A-B)+\\cos (A+B)}{\\cos (A-B)-\\cos (A+B)}=\\frac{-\\cos (C+D)+\\cos (C-D)}{-cos(C+D)-\\cos (C-D)}$<\/p>\n\n\n\n

$\\frac{2 \\cos A \\cos B}{2 \\sin A \\sin B}=\\frac{2 \\sin C \\sin D}{-2 \\cos C \\cos D}$<\/p>\n\n\n\n

cotA.cotB=-tanC.tanD<\/p>\n\n\n\n

$-1=\\frac{\\tan C \\tan B}{\\cot A \\cot B}$<\/p>\n\n\n\n

-1=tanA.tanB.tanC.tanD<\/p>\n\n\n\n

Question 25<\/h4>\n\n\n\n

Sol :
(i)
$\\tan (\\theta+\\phi)=3 \\tan \\theta$<\/p>\n\n\n\n

$\\frac{\\tan (\\theta+\\phi)}{\\tan \\theta}=\\frac{3}{1}$<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

$\\frac{\\tan (\\theta+\\phi)-\\tan \\theta}{\\tan (\\theta+\\phi)+\\tan \\theta}=\\frac{3-1}{3+1}$<\/p>\n\n\n\n

$\\dfrac{\\frac{\\sin (\\theta+\\phi)}{\\cos (\\theta+\\phi)}-\\frac{\\sin \\theta}{\\cos \\theta}}{\\frac{\\sin (\\theta+\\phi)}{\\cos (\\theta+\\phi)}+\\frac{\\sin \\theta}{\\cos \\theta}}=\\frac{2}{4}$<\/p>\n\n\n\n

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

$\\dfrac{\\sin \\left[\\theta+\\phi-\\theta\\right]}{\\sin [\\theta+\\phi+\\theta]}=\\frac{1}{2}$<\/p>\n\n\n\n

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

$\\sin (2 \\theta+\\phi)=2 \\sin \\phi$<\/p>\n\n\n\n

(ii)
$\\sin (2 \\theta+\\phi)=2 \\sin \\phi$<\/p>\n\n\n\n

[]<\/p>\n\n\n\n

[\u2235 sin2A=2sinA.cosA]<\/p>\n\n\n\n

$\\sin (2 \\theta+\\phi) \\cdot \\cos \\phi=2 sin \\phi cos\\phi$<\/p>\n\n\n\n

$\\frac{1}{2} \\times[2 \\sin (2 \\theta+\\phi) \\cos \\phi]=\\sin 2 \\phi$<\/p>\n\n\n\n

$\\frac{1}{2}[\\sin (2 \\theta+\\phi+\\phi)+\\sin (2 \\theta+\\phi-\\phi)]=\\sin 2 \\phi$<\/p>\n\n\n\n

$\\sin (2 \\theta+2 \\phi)+\\sin 2 \\theta=2 \\sin 2 \\phi$<\/p>\n\n\n\n

$\\sin 2(\\theta+\\phi)+\\sin 2 \\theta=2 \\sin 2 \\phi$<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 11 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0Sol :L.H.S=sin65\u00b0+cos65\u00b0=sin65\u00b0+sin(90\u00b0-65\u00b0)=sin65\u00b0+sin25\u00b0 \u2235$\\left[\\begin{array}{c} \\sin C+\\sin D =2 \\sin \\frac{C+D}{2} \\cos \\frac{C-D}{2}\\end{array}\\right]$ \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) $\\frac{\\cos 10^{\\circ}-\\sin 10^{\\circ}}{\\cos 10^{\\circ}+\\sin 10^{\\circ}}$Sol :L.H.S=$\\frac{\\cos 10^{\\circ}-\\sin 10^{\\circ}}{\\cos 10^{\\circ}+\\sin 10^{\\circ}}$ $=\\frac{\\sin (90^{\\circ}-10^{\\circ})-\\sin 10^{\\circ}}{\\sin (90^{\\circ}-10^{\\circ})+\\sin 10^{\\circ}}$ \u2235$\\left[\\begin{array}{c} \\sin C-\\sin D =2 cos \\frac{C+D}{2} sin \\frac{C-D}{2}\\end{array}\\right]$ \u2235$\\left[\\begin{array}{c} \\sin C+\\sin D =2 \\sin \\frac{C+D}{2} \\cos \\frac{C-D}{2}\\end{array}\\right]$  $=\\frac{2 […]<\/p>\n","protected":false},"author":302,"featured_media":625834,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[919],"tags":[],"boards":[],"class_list":["post-625827","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-11","entry"],"yoast_head":"\nKC Sinha: Exercise 6.2 - Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0Sol :L.H.S=sin65\u00b0+cos65\u00b0=sin65\u00b0+sin(90\u00b0-65\u00b0)=sin65\u00b0+sin25\u00b0 \u2235$left$ \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 6.2 - Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928\" \/>\n<meta property=\"og:description\" content=\"(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0Sol :L.H.S=sin65\u00b0+cos65\u00b0=sin65\u00b0+sin(90\u00b0-65\u00b0)=sin65\u00b0+sin25\u00b0 \u2235$left$ \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-08T06:38:28+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-08T06:38:34+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"8 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 6.2 – Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928\",\"datePublished\":\"2023-09-08T06:38:28+00:00\",\"dateModified\":\"2023-09-08T06:38:34+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/\"},\"wordCount\":2880,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg\",\"articleSection\":[\"class 11\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/\",\"name\":\"KC Sinha: Exercise 6.2 - 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Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928 - IndCareer Schools","description":"(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0Sol :L.H.S=sin65\u00b0+cos65\u00b0=sin65\u00b0+sin(90\u00b0-65\u00b0)=sin65\u00b0+sin25\u00b0 \u2235$left$ \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 6.2 - Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928","og_description":"(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0Sol :L.H.S=sin65\u00b0+cos65\u00b0=sin65\u00b0+sin(90\u00b0-65\u00b0)=sin65\u00b0+sin25\u00b0 \u2235$left$ \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-08T06:38:28+00:00","article_modified_time":"2023-09-08T06:38:34+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"8 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 6.2 – Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928","datePublished":"2023-09-08T06:38:28+00:00","dateModified":"2023-09-08T06:38:34+00:00","mainEntityOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/"},"wordCount":2880,"publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg","articleSection":["class 11"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/","url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/","name":"KC Sinha: Exercise 6.2 - Mathematics Solution Class 11 Chapter 6 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u0940\u092f \u092b\u0932\u0928 - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg","datePublished":"2023-09-08T06:38:28+00:00","dateModified":"2023-09-08T06:38:34+00:00","description":"(i) \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove that) sin65\u00b0+cos65\u00b0=\u221a2cos20\u00b0Sol :L.H.S=sin65\u00b0+cos65\u00b0=sin65\u00b0+sin(90\u00b0-65\u00b0)=sin65\u00b0+sin25\u00b0 \u2235$\\left$ \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f (Prove","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-2-mathematics-solution-class-11-chapter-6-trikonmitiy-phalan\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-7-2-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 6.4 - 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