Class 11 Maths Exercise 3.3 Solutions

L. H. S. \( =\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4} \)
\(=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}\)
\(=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}\)
\(=\text { R.H.S }\)
Hence, Proved.

L. H. S. \( =2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3} \)
\(=2\left(\frac{1}{2}\right)^{2}+\operatorname{cosec}^{2}\left\{\pi+\frac{\pi}{6}\right\}\left(\frac{1}{2}\right)^{2}\)
\(=2 \times \frac{1}{4}+\left\{-\operatorname{cosec} \frac{\pi}{6}\right\}^{2}\left(\frac{1}{4}\right)\)
\(=\frac{1}{2}+(-2)^{2}\left(\frac{1}{4}\right)\)
\(=\frac{1}{2}+\frac{4}{4}=\frac{1}{2}+1=\frac{3}{2}\)
\(=\text { RHS }\)
Hence, Proved

L. H. S. \( =\cot ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6} \)
\(=(\sqrt{3})^{2}+\operatorname{cosec}\left\{\pi-\frac{\pi}{6}\right\}+3\left(\frac{1}{\sqrt{3}}\right)^{2}\)
\(=3+\operatorname{cosec} {\frac{\pi}{6}}+3 \times \frac{1}{3}\)
\(=3+2+1=6\)
\(=\text { RHS }\)
Hence, proved.

L. H. S. \( =2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3} \)
\(=2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}\)
\(=2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8\)
\(=1+1+8\)
\(=10\)
\(=\text { RHS }\)
Hence, Proved.

\(\text { (i) } \sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)\)
\(=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}\)
\({[\sin (x+y)=\sin x \cos y+\cos x \sin y]}\)
\(=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
\(=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
(ii) \( \tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right) \)
\(=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}\quad {\left[\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right]}\)
\(=\frac{1-\frac{1}{\sqrt{3}}}{1+1\left(\frac{1}{\sqrt{3}}\right)}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2 \sqrt{3}}{\sqrt{3}^{2}-(1)^{2}}\)
\(=\frac{4-2 \sqrt{3}}{3-1}=2-\sqrt{3}\)

\(=\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\)
\(=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\right]\)
\(=\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}+\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]+\)
\(\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}-\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]\)
\(\quad[\therefore 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]\)
\(\quad[\therefore-2 \sin A \sin B=\cos (A+B)-\cos (A-B)]\)
\(=2 \times \frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]\)
\(=\cos \left[\frac{\pi}{2}-(x-y)\right]\)
\(=\sin (x+y)\)
\(=\text { R.H.S. }\)
Hence, proved.

It is known that
\( \tan (\mathrm{A}+\mathrm{B})=\frac{\tan A+\tan B}{1-\tan A \tan B} \) and \( \tan (\mathrm{A}-\mathrm{B})=\frac{\tan A-\tan B}{1+\tan A \tan B} \)
L.H.S. \( =\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\frac{\left(\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}\right)}{\left(\frac{\tan \frac{\pi}{4} \tan x}{1+\tan \frac{\pi}{4} \operatorname{tanx}}\right)} \)
\( =\frac{\left(\frac{1+\tan x}{1-\tan x}\right)}{\left(\frac{1-\operatorname{tanx}}{1-\tan x}\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}\)
\( = \) RHS
Hence, proved.

L.H.S. \( =\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x \)
\( =\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)} \)
\( =\frac{-\cos ^{2} x}{-\sin ^{2} x} \)
\( =\cot ^{2} x \)
\( = \) RHS
Hence, Proved.

L.H.S. \( =\cos \left(\frac{\pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{\pi}{x}-x\right)+\cot (2 \pi+x)\right] \)
\( =\sin x \cos x\left[\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi+x)\right] \)
\( [\therefore \) cos is positive in first and fourth quadrant]
\( =\sin x \cos x[\tan x+\cot x][\therefore \cot \) is positive in first and third quadrant \( ] \)
\( =\sin x \cos x\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right] \)
\( =\sin x \cos x\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right] \)
\( =\sin ^{2} x+\cos ^{2} x=1 \)
\( = \) RHS
Hence, proved.

\(\begin{array}{l}
\text { L.H.S. }=\sin (\mathrm{n}+1) x \sin (\mathrm{n}+2) x+\cos (\mathrm{n}+1) x \cos (\mathrm{n}+2) x \\
=\cos [(n+2) x-(n+1) x][\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})] \\
=\cos [n x+2 x-n x-x]=\cos x
\end{array}\)
R.H.S
Hence, proved

\(\text { L.H.S. }=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\)
\(=-2 \sin \left[\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right] \sin \left[\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right]\)
\(\because \left[\cos \mathrm{A}-\cos \mathrm{B}=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}\right]\)
\(=-2 \sin \left(\frac{3 \pi}{4}\right) \sin x\)
\(=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x\)
\(=-2 \sin \left(\frac{\pi}{4}\right) \sin x \quad [\because \sin \text { is positive in second quadrant] }\)
\(=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x\)
\(=-2 \sin \left(\frac{\pi}{4}\right) \sin x\quad [\sin \text { is positive in second quadrant] }\)
\(=-2 \times \frac{1}{\sqrt{2}} \times \sin x\)
\(=-\sqrt{2} \sin x\)
\(=\text { RHS }\)
Hence, proved

\(\text { L.H.S. }=\sin ^{2} 6 x-\sin ^{2} 4 x\)
\(=(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x)\)
\(=\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right]\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \cdot \sin \left(\frac{6 x-4 x}{2}\right)\right]\)
\(\quad \quad\left[\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \sin A-\sin B=\right.\)
\(\left.2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]\)
\(=(2 \sin 5 x \cos x)(2 \cos 5 x \sin x)\)
\(=(2 \sin 5 x \cos 5 x)(2 \sin x \cos x)\)
\(=\sin 10 x \sin 2 x\)
\(=\text { R.H.S. }\)
Hence, proved

\(\text { L.H.S. }=\cos ^{2} 2 x-\cos ^{2} 6 x\)
\(=(\cos 2 x+\cos 6 x)(\cos 2 x-6 x)\)
\(=\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \left(\frac{2 x-6 x}{2}\right)\right]\)
\(=[2 \cos 4 x \cos (-2 x)][-2 \sin 4 x \sin (-2 x)]\)
\(=(2 \cos 4 x \cos 2 x)[-2 \sin 4 x(-\sin 2 x)]\)
\(=(2 \sin 4 x \cos 4 x)(2 \sin 2 x \cos 2 x)\)
\(=\sin 8 x \sin 4x\)
\(=\text { RHS }\)
Hence, proved.

To prove: \( \sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos 2 x \sin 4 x \)
\(=\left[2 \sin \left(\frac{2 x+6 x}{2}\right)\left(\frac{2 x-6 x}{2}\right)\right]+2 \sin 4 x\)
\(=\left[\therefore \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]\)
\(=2 \sin 4 x \cos (-2 x)+2 \sin 4 x\)
\(=2 \sin 4 x \cos 2 x+2 \sin 4 x\)
\(=2 \sin 4 x(\cos 2 x+1)\)
\(=2 \sin 4 x(2 \cos 2 x-1+1)\)
\(=2 \sin 4 x(2 \cos 2 x)\)
\(=4 \cos 2 x \sin 4 x\)
\(=\text { R.H.S. }\)
Hence, proved.

\(\text { LHS }=\cot 4 x(\sin 5 x+\sin 3 x)\)
\(=\frac{\cot 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]\)
\(=\left[\therefore \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]\)
\(=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]\)
\( \mathrm{LHS}=2 \cos 4 x \cos x \)
\( \mathrm{RHS}=\cot x(\sin 5 x-\sin 3 x) \)
\(=\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \text { sin }\left(\frac{5 x-3 x}{2}\right)\right]\)
\(=\left[\therefore \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]\)
\(=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]\)
\( \mathrm{RHS}=2 \cos 4 x \cos x \)
\( \therefore \mathrm{LHS}=\mathrm{RHS} \)
Hence, proved

It is known that
\(=\cos \mathrm{A}-\cos \mathrm{B}=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \sin \mathrm{A}-\sin \mathrm{B}=2 \cos \left(\frac{A+B}{2}\right)\)
\(\sin \left(\frac{A-B}{2}\right)\)
L.H.S. \( =\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x} \)
\( =\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \cdot \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 x+3 x}{2}\right) \cdot \sin \left(\frac{17 x-3 x}{2}\right)} \)
\( =\frac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin n x} \)
\( =-\frac{\sin 2 x}{\cos 10 x} \)
\( \therefore \) LHS \( = \) RHS
Hence, proved.

It is known that
\( =\sin \mathrm{A}+\sin \mathrm{B}=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \cos \mathrm{A}+\cos \mathrm{B}=2 \cos \left(\frac{A+B}{2}\right) \) \( \cos \left(\frac{A-B}{2}\right) \)
L.H.S. \( =\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x} \)
\( =\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)} \)
\( =\frac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x} \)
\( =\frac{\sin 4 x}{\cos 4 x} \)
\( =\tan 4 x \)
\( \therefore \) LHS \( = \) RHS
Hence, proved.

L.H.S. \( =\frac{\sin x-\sin y}{\cos x+\cos y} \)
\( =\frac{2 \cos \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)} \)
\( =\frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)} \)
\( =\tan \left(\frac{x-y}{2}\right) \)
\( \therefore \) LHS \( = \) RHS
Hence, proved.

L.H.S. \( =\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x} \)
\( =\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)} \)
\( =\frac{\sin 2 x}{\cos 2 x} \)
\( =\tan 2 x \)
\( \therefore \) LHS \( = \) RHS
Hence, proved

L.H.S. \( =\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x \)
\(=\cot x \cot 2 x-\cot 3 x(\cot 2 x+\cot x)\)
\(=\cot x \cot 2 x-\cot (2 x+x)(\cot 2 x+\cot x)\)
\(=\cot x \cot 2 x-\left[\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}\right](\cot 2 x+\cot x)\)
\(=\left[\therefore \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}\right]\)
\(=\cot x \cot 2 x-(\cot 2 x \cot x-1)=1=\text { R.H.S. }\)

\(\text { L.H.S. }=\cos 4 x\)
\(=\cos 2(2 x)\)
\(=1-2 \sin ^{2} 2 x\left[\cos 2 \mathrm{~A}=1-2 \sin ^{2} \mathrm{~A}\right]\)
\(=1-2(2 \sin x \cos x) 2[\sin 2 \mathrm{~A}=2 \sin \mathrm{A} \operatorname{Cos} \mathrm{A}]\)
\(=1-8 \sin ^{2} x \cos 2 x=\text { R.H.S. }\)

\(\text { L.H.S. }=\cos 6 x\)
\(=\cos 3(2 x)\)
\(=4 \cos ^{3} 2 x-3 \cos 2 x\left[\cos 3 \mathrm{~A}=4 \cos ^{3} \mathrm{~A}-3 \cos \mathrm{A}\right]\)
\(=4\left[\left(2 \cos ^{2} x-1\right)^{3}-3\left(2 \cos ^{2} x-1\right)\left[\cos 2 x=2 \cos ^{2} x-1\right]\right.\)
\(=4\left[\left(2 \cos ^{2} x\right)^{3}-(1)^{3}-3\left(2 \cos ^{2} x\right)^{2}+3\left(2 \cos ^{2} x\right)\right]-6 \cos ^{2} x+3\)
\(=4\left[8 \cos ^{6} x-1-12 \cos ^{4} x+6 \cos ^{2} x\right]-6 \cos ^{2} x+3\)
\(=32 \cos ^{6} x-4-48 \cos ^{4} x+24 \cos ^{2} x-6 \cos ^{2} x+3\)
\(=32 \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1=\text { R.H.S. }\)