NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.3

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Chapter 3 Ex.3.3 Question 1

Prove that:

\({\sin ^2}\frac{\pi }{6} + {\cos ^2}\frac{\pi }{3} - {\tan ^2}\frac{\pi }{4} = - \frac{1}{2}\)

Solution

\[\begin{align}LHS &= {\sin ^2}\frac{\pi }{6} + {\cos ^2}\frac{\pi }{3} - {\tan ^2}\frac{\pi }{4}\\&= {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^2} - {\left( 1 \right)^2}\\&= \frac{1}{4} + \frac{1}{4} - 1\\&= \frac{{1 + 1 - 4}}{4}\\&= \frac{{ - 2}}{4}\\&= - \frac{1}{2} = RHS\end{align}\]

Chapter 3 Ex.3.3 Question 2

Prove that:

\(2{\sin ^2}\frac{\pi }{6} + {\rm{cose}}{{\rm{c}}^2}\frac{{7\pi }}{6}{\cos ^2}\frac{\pi }{3} = \frac{3}{2}\)

Solution

\[\begin{align}LHS &= 2{\sin ^2}\frac{\pi }{6} + {\rm{cose}}{{\rm{c}}^2}\frac{{7\pi }}{6}{\cos ^2}\frac{\pi }{3}\\&= 2{\left( {\frac{1}{2}} \right)^2} + {\rm{cose}}{{\rm{c}}^2}\left( {\pi + \frac{\pi }{6}} \right) \times {\left( \frac{1}{2} \right)^2}\\&= 2 \times \frac{1}{4} + {\left( { - {\rm{cosec}}\frac{\pi }{6}} \right)^2} \times \frac{1}{4}\\&= \frac{1}{2} + {\left( { - 2} \right)^2} \times \frac{1}{4}\\&= \frac{1}{2} + 1\\&= \frac{{1 + 2}}{2}\\&= \frac{3}{2} = RHS\end{align}\]

Chapter 3 Ex.3.3 Question 3

Prove that:

\({\cot ^2}\frac{\pi }{6} + {\rm{cose}}{{\rm{c}}^2}\frac{{5\pi }}{6} + 3{\tan ^2}\frac{\pi }{6} = 6\)

Solution

\[\begin{align}LHS &= {\cot ^2}\frac{\pi }{6} + {\rm{cosec}}\frac{{5\pi }}{6} + 3{\tan ^2}\frac{\pi }{6}\\&= {\left( {\sqrt 3 } \right)^2} + {\rm{cosec}}\left( {\pi - \frac{\pi }{6}} \right) + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2}\\&= 3 + {\rm{cosec}}\frac{\pi }{6} + 3 \times \frac{1}{3}\\&= 3 + 2 + 1\\&= 6 = RHS\end{align}\]

Chapter 3 Ex.3.3 Question 4

Prove that:

\(2{\sin ^2}\frac{{3\pi }}{4} + 2{\cos ^2}\frac{\pi }{4} + 2{\sec ^2}\frac{\pi }{3} = 10\)

Solution

\[\begin{align}LHS &= 2{\sin ^2}\frac{{3\pi }}{4} + 2{\cos ^2}\frac{\pi }{4} + 2{\sec ^2}\frac{\pi }{3}\\&= 2{\sin ^2}\left( {\pi - \frac{\pi }{4}} \right) + 2 \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times {\left( 2 \right)^2}\\&= 2{\sin ^2}\frac{\pi }{4} + 2 \times \frac{1}{2} + 2 \times 4\\&= 2 \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 1 + 8\\&= 1 + 9\\&= 10 = RHS \end{align}\]

Chapter 3 Ex.3.3 Question 5

Find the value of:

(i) \(\sin 75^\circ \)

(ii) \(\tan 15^\circ \)

Solution

(i) \(\sin 75^\circ \)

\[\begin{align} \sin 75{}^\circ&=\sin \left( 45{}^\circ +30{}^\circ \right) \\& =\sin 45{}^\circ \cos 30{}^\circ +\cos 45{}^\circ \sin 30{}^\circ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sin \left( x+y \right)=\sin x\cos y+\cos x\sin y \right] \\& =\left( \frac{1}{\sqrt{2}} \right)\times \left( \frac{\sqrt{3}}{2} \right)+\left( \frac{1}{\sqrt{2}} \right)\times \left( \frac{1}{2} \right) \\& =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\& =\frac{\sqrt{3}+1}{2\sqrt{2}}\end{align}\]

(ii) \(\tan 15^\circ \)

\[\begin{align} \tan 15{}^\circ &=\tan \left( 45{}^\circ -30{}^\circ \right) \\& =\frac{\tan 45{}^\circ -\tan 30{}^\circ }{1+\tan 45{}^\circ \tan 30{}^\circ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \tan \left( x-y \right)=\frac{\tan x-\tan y}{1+\tan x\tan y} \right] \\& =\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}} \\& =\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \\& =\frac{\sqrt{3}-1}{\sqrt{3}+1} \\& =\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \text{By}\ \text{rationalizing} \right] \\& =\frac{{{\left( \sqrt{3}-1 \right)}^{2}}}{3-1} \\& =\frac{3+1-2\sqrt{3}}{2} \\& =\frac{2\left( 2-\sqrt{3} \right)}{2} \\& =2-\sqrt{3}\end{align}\]

Chapter 3 Ex.3.3 Question 6

Prove the following:

\(\cos \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - y} \right) - \sin \left( {\frac{\pi }{4} - x} \right)\sin \left( {\frac{\pi }{4} - y} \right) = \sin \left( {x + y} \right)\)

Solution

\[\begin{align}LHS &= \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]\\ & = \left( \begin{array}{l}\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]\\\qquad + \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] \end{array} \right)\\\\&\qquad \quad\left[ \begin{array}{l}2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\\ - 2\sin A\sin B = \cos \left( {A + B} \right) - \cos \left( {A - B} \right)\end{array} \right]\\\\ &= \sin \left( {x + y} \right) = RHS\\ &= \left( \begin{array}{l}\frac{1}{2}\left[ {\cos \left\{ {\frac{\pi }{4} - x + \frac{\pi }{4} - y} \right\} + \cos \left\{ {\frac{\pi }{4} - x - \frac{\pi }{4} + y} \right\}} \right]\\\qquad + \frac{1}{2}\left[ {\cos \left\{ {\frac{\pi }{4} - x + \frac{\pi }{4} - y} \right\} - \cos \left\{ {\frac{\pi }{4} - x - \frac{\pi }{4} + y} \right\}} \right] \end{array} \right)\\ &= \frac{1}{2}\left[ {\cos \left\{ {\frac{\pi }{2} - \left( {x + y} \right)} \right\} + \cos \left\{ { - \left( {x - y} \right)} \right\} + \cos \left\{ {\frac{\pi }{2} - \left( {x + y} \right)} \right\} - \cos \left\{ { - \left( {x - y} \right)} \right\}} \right]\\& = \frac{1}{2}\left[ {2\cos \left\{ {\frac{\pi }{2} - \left( {x + y} \right)} \right\}} \right]\\& = \sin \left( {x + y} \right)\qquad \left[ {\cos \left( {\frac{\pi }{2} - A} \right) = \sin A} \right]\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 7

Prove the following:

\(\begin{align}\frac{\tan \left( {\frac{\pi }{4} + x} \right)}{\tan \left( {\frac{\pi }{4} - x} \right)} = \left( {\frac{1 + \tan x}{1 - \tan x}} \right)^2\end{align}\)

Solution

\[\begin{align}LHS& = \frac{{\tan \left( {\frac{\pi }{4} + x} \right)}}{{\tan \left( {\frac{\pi }{4} - x} \right)}}\\ & = \frac{{\left( {\frac{{\tan \frac{\pi }{4} + \tan x}}{{1 - \tan \frac{\pi }{4}\tan x}}} \right)}}{{\left( {\frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}}} \right)}}\qquad\left[ \begin{array}{l} \because\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\\\& \tan \left( {A - B} \right) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\end{array} \right]\\& = \frac{{\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)}}{{\left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right)}}\\ &= \left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right) \times \left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)\\ &= {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 8

Prove the following:

\(\begin{align}\frac{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}{\sin \left( {\pi - x} \right)\cos \left( {\frac{\pi }{2} + x} \right)} = {\cot ^2}x \end{align}\)

Solution

\[\begin{align}LHS &= \frac{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}{\sin \left( {\pi - x} \right)\cos \left( {\frac{\pi }{2} + x} \right)}\\ &= \frac{\left( { - \cos x} \right) \times \left( {\cos x} \right)}{\left( {\sin x} \right) \times \left( { - \sin x} \right)} \qquad \left[ \begin{array}{l}\because\cos \left( {\pi + x} \right) = - \cos x\\ \Rightarrow \cos \left(- x\right) = \cos x\\ \Rightarrow \cos \left(\frac{\pi }{2} + x\right) = - \sin x\\ \Rightarrow \sin \left(\pi - x\right) = \sin x\end{array} \right]\\ &= \frac{ - {{\cos }^2}x}{ - {{\sin }^2}x}\\ &= {\left(\frac{{\cos x}}{{\sin x}} \right)^2}\\& = {\cot ^2}x \qquad \left[\cot x = \frac{{\cos x}}{{\sin x}} \right]\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 9

Prove the following:

\(\cos \left( {\frac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\frac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] = 1\)

Solution

\[\begin{align}LHS &= \cos \left( {\frac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\frac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right]\\ &= \cos \left\{ {\pi + \left( {\frac{\pi }{2} + x} \right)} \right\}\cos x\left[ {\cot \left\{ {\pi + \left( {\frac{\pi }{2} - x} \right)} \right\} + \cot x} \right] \\& \qquad \quad \left[ \begin{array}{l}\because\cos \left( {2n\pi + \theta } \right) = \cos \theta \\ \Rightarrow \cot \left( {2n\pi + \theta } \right) = \cot \theta \end{array} \right]\\ &= - \cos \left( {\frac{\pi }{2} + x} \right)\cos x\left[ {\cot \left( {\frac{\pi }{2} - x} \right) + \cot x} \right]\\& \qquad \quad \left[ \begin{array}{l}\because\cos \left( {\pi + \theta } \right) = - \cos \theta \\ \Rightarrow \cot \left( {\pi + \theta } \right) = \cot \theta \end{array} \right]\\& = - \left( { - \sin x} \right)\cos x\left[ {\tan x + \cot x} \right]\\& \quad\qquad\left[ \begin{array}{l}\because\cos \left( {\frac{\pi }{2} + \theta } \right) = - \sin \theta \\ \Rightarrow \cot \left( {2n\pi + \theta } \right) = \cos \theta \end{array} \right]\\ & = \sin x\cos x\left[ {\frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}}} \right]\\ &= {\sin ^2}x + {\cos ^2}x\\& = 1\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 10

Prove the following:

\(\sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x = \cos x\)

Solution

\[\begin{align} LHS&=\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x \\ & =\cos \left( n+2 \right)x.\cos \left( n+1 \right)x+\sin \left( n+2 \right)x.\sin \left( n+1 \right)x \\ & =\cos \left\{ \left( n+2 \right)x-\left( n+1 \right)x \right\}\\& \qquad \quad \left[ \because \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \right] \\ & =\cos \left\{ n+2-n-1 \right\}x \\ & =\cos x \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 11

Prove the following:

\(\cos \left( {\frac{{3\pi }}{4} + x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right) = - \sqrt 2 \sin x\)

Solution

\[\begin{align} LHS&=\cos \left( \frac{3\pi }{4}+x \right)-\cos \left( \frac{3\pi }{4}-x \right) \\ & =-2\sin \left( \frac{\left( \frac{3\pi }{4}+x \right)+\left( \frac{3\pi }{4}-x \right)}{2} \right)\sin \left( \frac{\left( \frac{3\pi }{4}+x \right)-\left( \frac{3\pi }{4}-x \right)}{2} \right) \\ & \qquad \qquad \left[ \because \cos A-\cos B=-2\sin \left( \frac{A+B}{2} \right)\sin \left( \frac{A-B}{2} \right) \right] \\ & =-2\sin \left( \frac{\frac{3\pi }{4}+x+\frac{3\pi }{4}-x}{2} \right)\sin \left( \frac{\frac{3\pi }{4}+x-\frac{3\pi }{4}+x}{2} \right) \\ & =-2\sin \left( \frac{3\pi }{4} \right)\sin \left( \frac{2x}{2} \right) \\ & =-2\sin \left( \pi -\frac{\pi }{4} \right)\sin x \\ & =-2\sin \frac{\pi }{4}\sin x \qquad \qquad  \left[ \because \sin \left( \pi -\theta \right)=\sin \theta \right] \\ & =-2\times \frac{1}{\sqrt{2}}\times \sin x \\ & =-\sqrt{2}\sin x \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 12

Prove the following:

\({\sin ^2}6x - {\sin ^2}4x = \sin 2x\sin 10x\)

Solution

\[\begin{align}LHS &= {\sin ^2}6x - {\sin ^2}4x\\ &= \left( {\sin 6x + \sin 4x} \right)\left( {\sin 6x - \sin 4x} \right)\\ &= \left[ {2\sin \left( {\frac{{6x + 4x}}{2}} \right)\cos \left( {\frac{{6x - 4x}}{2}} \right)} \right] \times \left[ {2\cos \left( {\frac{{6x + 4x}}{2}} \right)\sin \left( {\frac{{6x - 4x}}{2}} \right)} \right]\\&\qquad \qquad\left[ \begin{array}{l}\because\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\\\& \sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\end{array} \right]\\ &= \left[ {2\sin 5x\cos x} \right] \times \left[ {2\cos 5x\sin x} \right]\\ &= \left[ {2\sin 5x\cos 5x} \right] \times \left[ {2\sin x\cos x} \right]\\ &= \left[ {\sin \left( {5x + 5x} \right) + \sin \left( {5x - 5x} \right)} \right] \times \left[ {\sin \left( {x + x} \right) + \sin \left( {x - x} \right)} \right]\\&\qquad \qquad\left[ {\because2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]\\ &= \left[ {\sin 10x + \sin 0} \right] \times \left[ {\sin 2x + \sin 0} \right]\\ &= \left[ {\sin 10x + 0} \right] \times \left[ {\sin 2x + 0} \right]\\ &= \sin 2x\sin 10x\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 13

Prove the following:

\({\cos ^2}2x - {\cos ^2}6x = \sin 4x\sin 8x\)

Solution

\[\begin{align}LHS &= {\cos ^2}2x - {\cos ^2}6x\\ &= \left( {\cos 2x + \cos 6x} \right)\left( {\cos 2x - \cos 6x} \right)\qquad \left[ \because{{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right]\\ &= \left[ {2\cos \left( {\frac{{2x + 6x}}{2}} \right)\cos \left( {\frac{{2x - 6x}}{2}} \right)} \right] \times \left[ { - 2\sin \left( {\frac{{2x + 6x}}{2}} \right)\sin \left( {\frac{{2x - 6x}}{2}} \right)} \right]\\& \qquad\qquad \left[ \begin{array}{l}\because\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\\\& \cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right) \end{array} \right]\\& = \left[ {2\cos 4x\cos \left( { - 2x} \right)} \right] \times \left[ { - 2\sin 4x\sin \left( { - 2x} \right)} \right]\\ &= \left[ {2\cos 4x\cos 2x} \right] \times \left[ { - 2\sin 4x\left( { - \sin 2x} \right)} \right]\\& \qquad\qquad\left[ \begin{array}{l}\because\cos \left( { - \theta } \right) = \cos \theta \\ \& \sin \left( { - \theta } \right) = - \sin \theta \end{array} \right]\\ &= \left[ {2\cos 4x\cos 2x} \right] \times \left[ {2\sin 4x\sin 2x} \right]\\ &= \left[ {2\cos 4x\sin 4x} \right] \times \left[ {2\cos 2x\sin 2x} \right]\\ &= \left[ {\sin \left( {4x + 4x} \right) - \sin \left( {4x - 4x} \right)} \right] \times \left[ {\sin \left( {2x + 2x} \right) - \sin \left( {2x - 2x} \right)} \right]\\& \qquad \qquad\left[ {\because 2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)} \right]\\ &= \left[ {\sin 8x - \sin 0} \right] \times \left[ {\sin 4x - \sin 0} \right]\\ &= \left[ {\sin 8x - 0} \right] \times \left[ {\sin 4x - 0} \right]\\ &= \sin 4x\sin 8x\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 14

Prove the following:

\(\sin 2x + 2\sin 4x + \sin 6x = 4{\cos ^2}x\sin 4x\)

Solution

\[\begin{align} LHS&=\sin 2x+2\sin 4x+\sin 6x \\ & =\left[ \sin 2x+\sin 6x \right]+2\sin 4x \\ & =\left[ 2\sin \left( \frac{2x+6x}{2} \right)\cos \left( \frac{2x-6x}{2} \right) \right]+2\sin 4x\\&\qquad \quad \left[ \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ & =\left[ 2\sin 4x\cos \left( -2x \right) \right]+2\sin 4x \\ & =2\sin 4x\cos 2x+2\sin 4x \\ & =2\sin 4x\left( \cos 2x+1 \right) \\ & =2\sin 4x\left( 2{{\cos }^{2}}x-1+1 \right) \qquad \quad \left[ \because \cos 2x=2{{\cos }^{2}}x-1 \right] \\ & =2\sin 4x\left( 2{{\cos }^{2}}x \right) \\ & =4{{\cos }^{2}}x\sin 4x \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 15

Prove the following:

\(\cot 4x\left( {\sin 5x + \sin 3x} \right) = \cot x\left( {\sin 5x - \sin 3x} \right)\)

Solution

\[\begin{align} LHS&=\cot 4x\left( \sin 5x+\sin 3x \right) \\ & =\cot 4x\left[ 2\sin \left( \frac{5x+3x}{2} \right)\cos \left( \frac{5x-3x}{2} \right) \right]\\&\qquad \quad  \left[ \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ & =\frac{\cos 4x}{\sin 4x}\left[ 2\sin 4x\cos x \right] \qquad \quad  \left[ \because \cos 2x=2{{\cos }^{2}}x-1 \right] \\ & =2\cos 4x\cos x \\ & =2\cos 4x\cos x\times \frac{\sin x}{\sin x} \\ & =\frac{\cos x}{\sin x}\times \left[ 2\cos 4x\sin x \right] \\ & =\cot x\left[ \sin \left( 4x+x \right)-\sin \left( 4x-x \right) \right]\\&\qquad \quad \left[ \because 2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right) \right] \\ & =\cot x\left( \sin 5x-\sin 3x \right) \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 16

Prove the following:

\(\begin{align}\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = - \frac{\sin 2x}{\cos 10x} \end{align}\)

Solution

\[\begin{align}LHS &= \frac{{\cos 9x - \cos 5x}}{{\sin 17x - \sin 3x}}\\ &= \frac{{\left[ { - 2\sin \left( {\frac{{9x + 5x}}{2}} \right)\sin \left( {\frac{{9x - 5x}}{2}} \right)} \right]}}{{\left[ {2\cos \left( {\frac{{17x + 3x}}{2}} \right)\sin \left( {\frac{{17x - 3x}}{2}} \right)} \right]}}\\& \qquad \qquad \left[ \begin{array}{l} \because \cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\\ \& \sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right) \end{array} \right]\\& = \frac{{\left[ { - 2\sin 7x\sin 2x} \right]}}{{\left[ {2\cos 10x\sin 7x} \right]}}\\ &= - \frac{{\sin 2x}}{{\cos 10x}}\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 17

Prove the following:

\(\begin{align}\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x \end{align}\)

Solution

\[\begin{align}LHS &= \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}\\& = \frac{{\left[ {2\sin \left( \frac{{5x + 3x}}{2} \right)\cos \left( {\frac{5x - 3x}{2}} \right)} \right]}}{{\left[ {2\cos \left( {\frac{5x + 3x}{2}} \right)\cos \left( {\frac{5x - 3x}{2}} \right)} \right]}}\\& \qquad \qquad \left[ \begin{array}{l} \because \sin A + \sin B = 2\sin \left( {\frac{A + B}{2}} \right)\cos \left( {\frac{A - B}{2}} \right)\\\& \cos A + \cos B = 2\cos \left( {\frac{A + B}{2}} \right)\cos \left( {\frac{A - B}{2}} \right)\end{array} \right]\\& = \frac{\sin 4x}{\cos 4x}\\ &= \tan 4x\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 18

Prove the following:

\(\begin{align}\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2} \end{align}\)

Solution

\[\begin{align}LHS &= \frac{\sin x - \sin y}{\cos x + \cos y}\\ &= \frac{{\left[ {2\cos \left( {\frac{x + y}{2}} \right)\sin \left( {\frac{x - y}{2}} \right)} \right]}}{{\left[ {2\cos \left( {\frac{x + y}{2}} \right)\cos \left( {\frac{x - y}{2}} \right)} \right]}}\\& \qquad \qquad \left[ \begin{array}{l}\because \sin A - \sin B = 2\cos \left( {\frac{A + B}{2}} \right)\sin \left( {\frac{A - B}{2}} \right)\\\& \cos A + \cos B = 2\cos \left( {\frac{A + B}{2}} \right)\cos \left( {\frac{A - B}{2}} \right)\end{array} \right]\\ &= \frac{{\sin \left( {\frac{x - y}{2}} \right)}}{{\cos \left( {\frac{x - y}{2}} \right)}}\\ &= \tan \frac{x - y}{2}\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 19

Prove the following:

\(\begin{align}\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x \end{align}\)

Solution

\[\begin{align}LHS &= \frac{\sin x + \sin 3x}{\cos x + \cos 3x}\\& = \frac{{\left[ {2\sin \left( {\frac{x + 3x}{2}} \right)\cos \left( {\frac{x - 3x}{2}} \right)} \right]}}{{\left[ {2\cos \left( {\frac{x + 3x}{2}} \right)\cos \left( {\frac{x - 3x}{2}} \right)} \right]}}\\& \qquad \qquad \left[ \begin{array}{l}\because \sin A + \sin B = 2\sin \left( {\frac{A + B}{2}} \right)\cos \left( {\frac{A - B}{2}} \right)\\\& \cos A + \cos B = 2\cos \left( {\frac{A + B}{2}} \right)\cos \left( {\frac{A - B}{2}} \right)\end{array} \right]\\ &= \frac{\sin 2x}{\cos 2x}\\ &= \tan 2x\\& = RHS\end{align}\]

Chapter 3 Ex.3.3 Question 20

Prove the following:

\(\begin{align}\frac{\sin x - \sin 3x}{{{\sin }^2x - {\cos ^2}x}} = 2\sin x \end{align}\)

Solution

\[\begin{align}LHS &= \frac{\sin x - \sin 3x}{{{\sin }^2x - {\cos }^2}x}\\ &= \frac{{\left[ {2\cos \left( {\frac{x + 3x}{2}} \right)\sin \left( {\frac{x - 3x}{2}} \right)} \right]}}{{ - \left[ {{\cos }^2x - {\sin }^2}x \right]}}\\& \qquad \qquad \left[ \begin{array}{l}\because \sin A - \sin B = 2\cos \left( {\frac{A + B}{2}} \right)\sin \left( {\frac{A - B}{2}} \right)\\\& \cos A + \cos B = 2\cos \left( {\frac{A + B}{2}} \right)\cos \left( {\frac{A - B}{2}} \right) \end{array} \right]\\ &= \frac{2\cos 2x\sin \left(- x \right)}{- \cos 2x}\\ &= 2\sin x\\ &= RHS\end{align}\]

Chapter 3 Ex.3.3 Question 21

Prove the following:

\(\begin{align}\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x \end{align}\)

Solution

\[\begin{align}LHS &= \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}\\ &= \frac{{\left[ \cos 4x + \cos 2x \right] + \cos 3x}}{{\left[ \sin 4x + \sin 2x \right] + \sin 3x}}\\ &= \frac{{\left[ {2\cos \left( \frac{{4x + 2x}}{2} \right)\cos \left(\frac{4x - 2x}{2} \right)} \right] + \cos 3x}}{{\left[ {2\sin \left(\frac{4x + 2x}{2} \right)\cos \left(\frac{4x - 2x}{2} \right)} \right] + \sin 3x}} \\& \qquad \qquad \left[ \begin{array}{l}\because\cos A + \cos B = 2\cos \left(\frac{A + B}{2} \right)\cos \left(\frac{A - B}{2} \right)\\\& \sin A + \sin B = 2\sin \left(\frac{A + B}{2}\right)\cos \left(\frac{A - B}{2} \right) \end{array} \right]\\& = \frac{2\cos 3x\cos x + \cos 3x}{2\sin 3x\cos x + \sin 3x}\\ &= \frac{\cos 3x\left( {2\cos x + 1} \right)}{{\sin 3x\left( {2\cos x + 1} \right)}}\\ &= \frac{\cos 3x}{\sin 3x}\\& = \cot 3x\\& = RHS \end{align}\]

Chapter 3 Ex.3.3 Question 22

Prove the following:

\(\cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1\)

Solution

\[\begin{align} LHS&=\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x \\ & =\cot x\cot 2x-\cot 3x\left( \cot 2x+\cot x \right) \\ & =\cot x\cot 2x-\cot \left( 2x+x \right)\left( \cot 2x+\cot x \right) \\ & =\cot x\cot 2x-\left[ \frac{\cot 2x\cot x-1}{\cot 2x+\cot x} \right]\left( \cot 2x+\cot x \right)\\&\qquad \qquad \left[ \because \cot \left( A+B \right)=\frac{\cot A\cot B-1}{\cot A+\cot B} \right] \\ & =\cot x\cot 2x-\left[ \cot 2x\cot x-1 \right] \\ & =\cot x\cot 2x-\cot x\cot 2x+1 \\ & =1 \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 23

Prove the following:

\(\tan 4x = \frac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}\)

Solution

\[\begin{align} LHS&=\tan 4x \\ & =\tan 2\left( 2x \right) \\ & =\frac{2\tan 2x}{1-{{\tan }^{2}}2x}\\ & \qquad \qquad \left[ \because \tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A} \right] \\ & =\frac{2\left( \frac{2\tan x}{1-{{\tan }^{2}}x} \right)}{1-{{\left( \frac{2\tan x}{1-{{\tan }^{2}}x} \right)}^{2}}}\\ & \qquad \qquad \left[ \because \tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A} \right] \\ & =\frac{\left( \frac{4\tan x}{1-{{\tan }^{2}}x} \right)}{1-\left( \frac{4{{\tan }^{2}}x}{1+{{\tan }^{4}}x-2{{\tan }^{2}}x} \right)}\\ & \qquad \qquad  \left[ \because {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right] \\ & =\frac{\left( \frac{4\tan x}{1-{{\tan }^{2}}x} \right)}{\left( \frac{1+{{\tan }^{4}}x-2{{\tan }^{2}}x-4{{\tan }^{2}}x}{1+{{\tan }^{4}}x-2{{\tan }^{2}}x} \right)} \\ & =\left( \frac{4\tan x}{1-{{\tan }^{2}}x} \right)\times \left( \frac{1+{{\tan }^{4}}x-2{{\tan }^{2}}x}{1+{{\tan }^{4}}x-6{{\tan }^{2}}x} \right) \\ & =\frac{4\tan x{{\left( 1-{{\tan }^{2}}x \right)}^{2}}}{\left( 1-{{\tan }^{2}}x \right)\left( 1+{{\tan }^{4}}x-6{{\tan }^{2}}x \right)}\\ & \qquad \qquad  \left[ \because {{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}} \right] \\ & =\frac{4\tan x\left( 1-{{\tan }^{2}}x \right)}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x} \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 24

Prove the following:

\(\cos 4x = 1 - 8{\sin ^2}x{\cos ^2}x\)

Solution

\[\begin{align}LHS & =\cos 4x \\ & =\cos 2\left( 2x \right) \\ & =1-2{{\sin }^{2}}2x \\& \qquad \quad \left[ \because \cos 2A=1-2{{\sin }^{2}}x \right] \\ & =1-2{{\left( 2\sin x\cos x \right)}^{2}}\\& \qquad \quad \left[ \because \sin 2A=2\sin x\cos x \right] \\ & =1-2\left( 4{{\sin }^{2}}x{{\cos }^{2}}x \right) \\ & =1-8{{\sin }^{2}}x{{\cos }^{2}}x \\ & =RHS \end{align}\]

Chapter 3 Ex.3.3 Question 25

Prove the following:

\(\cos 6x = 32{\cos ^6}x - 48{\cos ^4}x + 18{\cos ^2}x - 1\)

Solution

\[\begin{align} LHS &=\cos 6x \\ & =\cos 3\left( 2x \right) \\ & =4{{\cos }^{3}}2x-3\cos 2x \\& \qquad \qquad \left[ \because \cos 3A=4{{\cos }^{3}}x-3\cos x \right] \\ & =4{{\left( 2{{\cos }^{2}}x-1 \right)}^{3}}-3\left( 2{{\cos }^{2}}x-1 \right)\\& \qquad \qquad \left[ \because \cos 2A=2{{\cos }^{2}}x-1 \right] \\ & =4\left( 8{{\cos }^{6}}x-1-12{{\cos }^{4}}x+6{{\cos }^{2}}x \right)-6{{\cos }^{2}}x+3\\& \qquad \qquad \left[ \because {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} \right] \\ & =32{{\cos }^{6}}x-4-48{{\cos }^{4}}x+24{{\cos }^{2}}x-6{{\cos }^{2}}x+3 \\ & =32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1 \\ & =RHS \end{align}\]