# NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.4

Go back to  'Introduction to Trigonometry'

## Chapter 8 Ex.8.4 Question 1

Express the trigonometric ratios $$\text{sin} \;A$$, $$\text{sec} \;A$$ and $$\text{ tan} \;A$$ in terms of $$\text{cot} \;A$$.

#### Reasoning:

\begin{align}{ \text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}

#### Steps:

Consider a  \begin{align}\Delta {ABC}\end{align} with \begin{align}\angle {B }=\text{ }90{}^\circ \end{align}

Using the Trigonometric Identity,

$$\text{cose}{{\text{c}}^{2}}\,{A}=1+{{\cot }^{2}}{A}$$

\begin{align} \frac{1}{\text{cose}{{\text{c}}^{2}}\,{A}}&=\frac{\text{1}}{\text{1+co}{{\text{t}}^{\text{2}}}{A}} \end{align} (By taking reciprocal both the sides)

\begin{align} &{{\text{sin}}^{2}}{A}=\frac{1}{1+{{\cot }^{2}}{A}} \\ & \left( \text{As }\frac{1}{\text{cose}{{\text{c}}^{2}}\,\text{A}}={{\sin }^{2}}\text{A} \right) \\ \end{align}

Therefore,

\begin{align}\text{sin}\,{A}\,=+\frac{1}{\sqrt{1+{{\cot }^{2}}{A}}} \end{align}

For any sine value with respect to an angle in a triangle, sine value will never be negative. Since, sine value will be negative for all angles greater than 180°.

Therefore, \begin{align}\text{sin}\,{A}\,=\frac{\text{1}}{\sqrt{\text{1+co}{{\text{t}}^{\text{2}}}{A}}} \end{align}

We know that,

\begin{align}\text{tan} A=\frac{\text{sin}} A{\cos \,{A}}\end{align}

However, Trigonometric Function,

\begin{align}\text{cot}\,{A=}\frac{\cos \,{A}}{\text{Sin}\,{A}}\end{align}

Therefore, Trigonometric Function,

\begin{align}\text{tan}\,A=\frac{1}{\cot \,{A}}\end{align}

Also,

\begin{align}\text{se}{{\text{c}}^{\text{2}}} A = 1 + {{\text{tan}}^{\text{2}}}{A}\end{align} (Trigonometric Identity)

\begin{align}=1+\frac{1}{{{\cos }^{2}}{A}} \\ =\frac{{{\cot }^{2}}\,{A+1}}{{{\cot }^{2}}{A}} \\ \end{align}

\begin{align}\text{sec}\,{A}=\frac{\sqrt{{{\cot }^{2}}{A+1}}}{\cot \,A}\end{align}

## Chapter 8 Ex.8.4 Question 2

Write all the other trigonometric ratios of$$\angle A$$ in terms of $$\text{sec} \;A$$.

#### Reasoning:

\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}

#### Steps:

We know that, Trigonometric Function,

\begin{align}\text{cos}\,{A = }\frac{1}{\sec {A}}\ldots \text{Equation }\left( \text{1} \right)\end{align}

Also,

\begin{align} & \text{si}{{\text{n}}^{\text{2}}}A + \text{co}{{\text{s}}^{\text{2}}}{A} = {1 } \\ & \left( \text{Trigonometric identity} \right) \\ \\& \text{si}{{\text{n}}^{\text{2}}}{A} = {1 }-\text{ co}{{\text{s}}^{\text{2}}}{A } \\ & \left( \text{By transposing} \right) \\ \end{align}

Using value of $$\text{cos} \;A$$ from Equation $$(1)$$ and simplifying further,

\begin{align}\sin {{A} }&=\sqrt {1 - {{\left( {\frac{1}{{\sec {{A}}}}} \right)}^2}} \\&= \sqrt {\frac{{{{\sec }^2}\,{{A}} - 1}}{{{{\sec }^2}\,{{A}}}}} \\ &= \frac{{\sqrt {{{\sec }^2}{{A}} - 1} }}{{\sec {{A}}}}\ldots (2)\end{align}

\begin{align} & {\text{ta}}{{\text{n}}^{\text{2}}}{{A + 1 }}={{\text{sec}}^{\text{2}}}{{A }} \\ & \left( {{\text{Trigonometric identity}}} \right)\\\\ & {\text{ta}}{{\text{n}}^{\text{2}}}{{A}}={{\text{sec}}^{\text{2}}}{{A - 1 }} \\ & \left( {{\text{By transposing}}} \right) \\\end{align}

Trigonometric Function,

\begin{align} {\text{tan}}\,{\text{A}}&=\sqrt {{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{1}}}\,\,\dots\left( {\text{3}} \right)\\ {\text{cot}}\,{\text{A}}\,& =\,\frac{{{\text{cosA}}}}{{{\text{sinA}}}} \\&= \frac{{\frac{1}{{\sec \,{\text{A}}}}}}{{\frac{{\sqrt {{{\sec }^2}} {\text{A}} - 1}}{{\sec \,{\text{A}}}}}}\\& \begin{bmatrix}\text{ (By substituting}\\ \text{equations (1) and (2)}\end{bmatrix}\\\\&= \frac{1}{{\sqrt {{{\sec }^2}{\text{A}} - {\text{1}}} }} \\{\rm{cosec}}\,{\text{A}}&=\frac{1}{{\sin {\text{A}}}}\\&= \frac{{\sec {\text{A}}}}{{\sqrt {{{\sec }^2}\,{\text{A}} - 1} }}\\\\&\begin{bmatrix}\text{(By substituting}\\\text{ Equation (2) and}\\\text{simplifying)}\end{bmatrix}\end{align}

## Chapter 8 Ex.8.4 Question 3

Evaluate

(i) $$\,\,\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}}$$

(ii) $$\,\,\,\text{sin}{{25}^{^\circ }}\text{cos}{{65}^{^\circ }}+\text{cos}{{25}^{^\circ }}\text{ sin}{{65}^{^\circ }}$$

#### Reasoning:

\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\sin \left(90^{\circ}-\theta\right)=\cos \theta} \\ {\cos \left(90^{\circ}-\theta\right)=\sin \theta}\end{align}

#### Steps:

(i)\begin{align}\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}}\end{align}

\begin{align} & =\frac{{{\left[ \sin \left( {{90}^{{}^\circ }}-27 \right) \right]}^{2}}+{{\sin }^{2}}27}{{{\left[ \cos \left( {{90}^{{}^\circ }}-{{73}^{{}^\circ }} \right) \right]}^{2}}+{{\cos }^{2}}{{73}^{{}^\circ }}} \\ & =\frac{{{[\cos 27]}^{2}}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\left[ \sin {{73}^{{}^\circ }} \right]}^{2}}+{{\cos }^{2}}{{73}^{{}^\circ }}} \\ & \begin{bmatrix} \sin \left( {{90}^{{}^\circ }}-\text{ }\!\!\theta\!\!\text{ } \right)=\cos \text{ }\!\!\theta\!\! \\ \And \\ \cos \left( {{90}^{{}^\circ }}-\theta \right)=\sin \theta \end{bmatrix} \\ & =\frac{1}{1} \; \begin{bmatrix} \text{By identity} \\ \sin ^{2} A+\cos ^{2} A=1 \end{bmatrix} \\& =1\end{align}

(ii) $$\sin {{25}^{{}^\circ }}\cos {{65}^{{}^\circ }}+\cos {{25}^{{}^\circ }}\sin {{65}^{{}^\circ }}$$

\begin{align} &= \left( \sin {{25}^{{}^\circ }} \right)\left\{ \cos \left( {{90}^{{}^\circ }}-{{25}^{{}^\circ }} \right) \right\} +\cos {{25}^{{}^\circ }} \left\{ \sin \left( {{90}^{{}^\circ }}-{{25}^{{}^\circ }} \right) \right. \\ & \begin{bmatrix} \sin \left( {{90}^{{}^\circ }}-\text{ }\!\!\theta\!\!\text{ } \right)=\cos \text{ }\!\!\theta\!\! \\ \And \cos \left( {{90}^{{}^\circ }}-\theta \right)=\sin \theta \end{bmatrix} \\ & = \begin{pmatrix} \left( \sin {{25}^{{}^\circ }} \right)\left( \sin {{25}^{{}^\circ }} \right) +\cos {{25}^{{}^\circ }}\left( \cos {{25}^{{}^\circ }} \right) \end{pmatrix}\\ & ={{\sin }^{2}}{{25}^{{}^\circ }}+{{\cos }^{2}}{{25}^{{}^\circ }} \\ & =1 \\ \\& \begin{bmatrix} \text{By identity} \\ \sin ^{2} A+\cos ^{2} A=1 \end{bmatrix} \end{align}

## Chapter 8 Ex.8.4 Question 4

Choose the correct option. Justify your choice.

(i)  $$\text{ 9 sec 2A - 9 tan 2A} = \_\_\_\_\_\_\_\_\_\_\_\_\_$$

(A) $$1$$

(B) $$9$$

(C) $$8$$

(D) $$0$$

(ii) $$\left( \text{1 + tan } \theta\text{ + sec }\theta \right) \left( \text{1 + cot } \theta -\text{cosec }\theta \right) = \_\_\_\_\_\_\_\_\_\_\_\_\_$$

(A) $$0$$

(B) $$1$$

(C) $$2$$

(D) $$-1$$

(iii) $$\left( \text{sec A + tan A} \right) \left( \text{1 -sin A} \right) =\_\_\_\_\_\_\_\_\_$$

(A) $$\text{ sec A }$$

(B) $$\text{ sin A }$$

(C) $$\text{ cosec A }$$

(D) $$\text{ cos A}$$

(iv) $$\frac{\text{1+ta}{{\text{n}}^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} = \_\_\_\_\_$$

(A) $$\text{ sec2A}$$

(B) $$1$$

(C) $$\text{cot2A}$$

(D) $$\text{tan2A}$$

#### Steps:

(i)

\begin{align} & \text{ 9 se}{{\text{c}}^{\text{2}}}\text{A }-\text{ 9 ta}{{\text{n}}^{\text{2}}}\text{A} \\ \\ &=9\text{ (se}{{\text{c}}^{\text{2}}}\text{A}-\text{ta}{{\text{n}}^{\text{2}}}\text{A) } \\& =\text{9 }\left( \text{1} \right) \\ & \begin{bmatrix} \text{By the identity,} \\ 1+ \sec ^2 A = \tan ^2 A \\ \Rightarrow \sec ^2 A - \tan^2A=1 \end{bmatrix} \\ & = 9\end{align}

(ii)

\begin{align} \left( \text{1 + tan }\theta\text{ + sec } \theta \right) \left( \text{1 + cot }\theta\text{ - cosec }\theta \right) \ldots (1) \end{align}

We know that the trigonometric functions,

\begin{align} & \tan (x)=\frac{\sin (x)}{\cos (x)} \\ & \cot (x)=\frac{\cos (x)}{\sin (x)}=\frac{1}{\tan (x)} \end{align}

And

\begin{align} \sec (x)&=\frac{1}{\cos (x)} \\ {cosec}(x)&=\frac{1}{\sin (x)} \end{align}

By substituting the above function in Equation $$(1)$$,

\begin{align}& = \begin{bmatrix} \left( 1+ \frac{\text{sin }\theta }{\text{cos }\theta } + \frac{1}{\text{cos }\theta } \right) \left( 1+ \frac{\text{cos }\theta }{\text{sin }\theta } - \frac{1}{\text{sin }\theta} \right) \end{bmatrix} \\ & = \begin{bmatrix} \left( \frac{\text{cos }\theta\text{ +sin }\theta+1}{\text{cos }\theta} \right) \left( \frac{\text{sin }\theta\text{ +cos }\theta -1}{\text{sin }\theta} \right)\end{bmatrix}\\& \text{(By taking LCM and multiplying)}\\\\& = \frac{{{\text{(sin }\theta +\text{cos} \theta )}^2}-(1)^2}{\text{sin }\theta \text{cos }\theta}\\& \text{(Using )}\\ \\&= \frac{ \begin{Bmatrix}\text{sin}^2 \theta \text{ +co}{{\text{s}}^2} \theta \\ \text{ +2sin }\theta\text{ cos }\theta -1 \end{Bmatrix} }{\text{sin }\theta\text{ cos }\theta} \\ & = \frac{\text{1+2sin }\theta\text{ cos }\theta -1 }{\text{sin }\theta\text{ cos}\theta}\\&\text{(Using identify)}\\\\&=\frac{\text{2sin }\theta\text{ cos }\theta\text{ }}{\text{sin }\theta\text{ cos }\theta\text{ }}\text{= 2}\end{align}

Hence, option (C) is correct.

(iii)

\begin{align}\left( \text{sec A + tan A} \right) \left( 1 - \text{sin A} \right) \ldots \left( 1 \right)\end{align}

We know that the trigonometric functions,

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

And

\begin{align}\sec (x)=\frac{1}{\cos (x)}\end{align}

By substituting the above function in Equation $$(1)$$,

\begin{align} &= \begin{bmatrix} \left( \frac{1}{\text{cosA}} + \frac{\text{sinA}}{\text{cosA}} \right) (1-\text{sinA)} \end{bmatrix} \\ & =\left( \frac{\text{1+sinA}}{\text{cosA}} \right) (1-\text{sinA)} \\ & = \frac{1-\text{si}{{\text{n}}^{2}}\text{A}}{\text{cosA}} \\ & = \frac{\text{co}{{\text{s}}^{2}}\text{A}}{\text{cosA}} \\ & \begin{bmatrix} \text{By identity}\\\sin^2 \theta+ \cos^{2} \theta=1 \\ \Rightarrow 1- {\rm{sin^2}} θ = {\rm{cos^2}} θ \end{bmatrix} \\ &= \cos A\end{align}
Hence, option (D) is correct.

(iv) \begin{align}\frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}}\end{align}

We know that the trigonometric functions,

\begin{align} \tan (x)&=\frac{\sin (x)}{\cos (x)} \\ \cot (x)&=\frac{\cos (x)}{\sin (x)} \\ & =\frac{1}{\tan (x)} \end{align}

By substituting the above function in Equation ($$1$$),

\begin{align}\frac{{{\rm{1 + ta}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{1 + co}}{{\rm{t}}^{{2}}}{{A}}}}&= \frac{{{{1 + }}\frac{{{\rm{sin}}^2{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{{{1 + }}\frac{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}{{{\rm{si}}{{\rm{n}}^{{2}}}{\rm{A}}}}}}\\ &= \frac{{\frac{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}} + {\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{\frac{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}} + {\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}}}\\&=\frac{{\frac{{{1}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{\frac{1}{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}}}\\&=\frac{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}\\&= \rm{tan}^2 A\end{align}

Hence, option ($$D$$) is correct

## Chapter 8 Ex.8.4 Question 5

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) \begin{align} {{\text{(cosec}\,\text{ }\!\!\theta\!\!\text{ }-\text{cot}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}\text{=}\frac{\text{1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ }}{\text{1+cos}\,\text{ }\!\!\theta\!\!\text{ }} \end{align}

(ii) \begin{align} \frac{\text{cos}\,A}{\text{1+sin}\,A}\,\,\text{+}\,\frac{\text{1+sin}\,A}{\text{cos}\,A}\text{=2sec}A \end{align}

(iii) \begin{align} \begin{bmatrix} \frac{\text{tan} \theta }{1-\text{cot} \theta} +\frac{\text{cot} \theta}{1-\text{tan} \theta}\\ =1+\text{sec} \theta \text{cosec} \theta \end{bmatrix} \end{align}

(iv) \begin{align} \frac{\text{1+sec}\; A}{\text{sec}\,A}\text{=}\frac{\text{si}{{\text{n}}^{\text{2}}}A}{\text{1}-\text{cos}\,A} \end{align}

(v) \begin{align} \begin{bmatrix} \frac{\text{cos}\,\text{A}-\text{sin}\,\text{A+1}}{\text{cos}\,\text{A+sin}\,\text{A+1}} \\ =\text{cosec}\,\text{A}\,\text{+}\,\text{cotA} \end{bmatrix} \end{align}

(vi) \begin{align} \sqrt{\frac{\text{1+sinA}}{\text{1-sinA}}}\,\text{=}\,\text{sec}\,\text{A}\,\text{+}\,\text{tan}\,\text{A} \end{align}

(vii) \begin{align} \frac{\text{sin}\,\text{ }\!\!\theta\!\!\text{ }-\text{2si}{{\text{n}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2cos}\,\text{ }\!\!\theta\!\!\text{ }-\text{cos}\,\text{ }\!\!\theta\!\!\text{ }}\,\text{=}\,\text{tan}\,\text{ }\!\!\theta\!\!\text{ } \end{align}

(viii) \begin{align} \begin{bmatrix} {{\text{(sin} A\,\text{+}\,\text{cosec}\,A)}^{\text{2}}}\\+(cos\,A\,\text{+}\,\text{sec}\,A{{\text{)}}^{\text{2}}} \\ =\text{7+ta}{{\text{n}}^{\text{2}}}A\,\text{+}\,\text{co}{{\text{t}}^{\text{2}}}A \end{bmatrix} \end{align}

(ix) \begin{align} \begin{bmatrix} \text{(cosecA}-\text{sinA)}\\\text{(secA}-\text{cosA)} \\ =\frac{\text{1}}{\,\text{tan}\,\text{A+cot}\,\text{A}} \end{bmatrix} \end{align}

(x) \begin{align} \begin{bmatrix} \left( \frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} \right) \\ ={{\left( \frac{1-\tan \text{A}}{1-\cot \text{A}} \right)}^{2}}\\ = \rm tan^2 A \end{bmatrix} \end{align}

#### Reasoning:

\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}

#### Steps:

(i)\begin{align} \;\; {(\text{cosec}\; \theta-\cot \theta)^{2}}&={\frac{1-\cos \theta}{1+\cos \theta}} \end{align}

\begin{align} {\text { L.H.S }}&={(\text{cosec}\; \theta-\cot \theta)^{2}} \\ &\qquad \quad \ldots (1) \end{align}

We know that the trigonometric functions,

\begin{align} \cot \,(x)&=\frac{\cos \,(x)}{\sin \,(x)} =\frac{1}{\tan \,(x)} \\ \operatorname{cosec}\,(x) &=\frac{1}{\sin \,(x)} \end{align}

By substituting the above function in Equation $$(1)$$

\begin{align} \rm (cosec \theta - cot \theta)^2 &\!\!= \!\!{{\left( \frac{1}{\text{sin } \theta }-\frac{\text{cos} \theta }{\text{sin} \theta} \right)}^{2}}\end{align}

\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{{{\text{(sin}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}\,\,\,\,\,\,\end{align}

\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{\text{si}{{\text{n}}^{\text{2}}}\,\text{ }\!\!\theta\!\!\text{ }} \end{align}

\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{\text{1}-\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}\,\,\,\,\,\,\,\, \end{align} (By identity $$\sin^2 A$$ $$+$$ $$\cos^2 A =$$ $$1$$ Hence, $$1-\cos^2 A$$$$= \sin^2 A$$)

\begin{align} \text{=}\,\frac{{{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}^{\text{2}}}}{\text{(1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ )(1+cos}\,\text{ }\!\!\theta\!\!\text{ )}} \end{align} [Using $$a^2 - b^2$$$$=$$$$(a+b)(a-b)$$]

\begin{align} \text{=}\,\frac{\text{1}-\text{cos}\,\text{ }\!\!\theta\!\!\text{ }}{\text{1+cos}\,\text{ }\!\!\theta\!\!\text{ }} \end{align}

=R.H.S.

(ii) \begin{align} \frac{\text{cos}A}{\text{1+sin}A}&+\frac{\text{1+sin}A}{\text{cos}A}\,\,=\,\text{2sec}A \end{align}

LHS\begin{align} =\,\frac{\text{cosA}}{\text{1+sinA}}\text{+}\frac{\text{1+sinA}}{\text{cosA}} \end{align}

\begin{align} \text{=}\,\frac{\text{co}{{\text{s}}^{\text{2}}}\text{A+(1+sinA}{{\text{)}}^{\text{2}}}}{\text{(1+sinA)}\,\text{(cosA)}} \end{align}

\begin{align} \text{=}\,\,\frac{\text{co}{{\text{s}}^{\text{2}}}A\text{+1+si}{{\text{n}}^{\text{2}}}A\text{+2sin}A}{\text{(1+sin}A)\,\text{(cos}A)} \end{align}

\begin{align} \text{=}\frac{\text{si}{{\text{n}}^{\text{2}}}A\text{+co}{{\text{s}}^{\text{2}}}A\text{+1+2sin}A}{\text{(1+sin}A)\text{(cos}A)} \end{align}

\begin{align} \text{=}\frac{\text{1+1+2sin}A}{\text{(1+sin}A)\text{(cos}A)} \end{align} (By identify $$\sin^2 A + \cos ^2 A=1$$)

\begin{align} \text{=}\frac{\text{2+2sin}A}{\text{(1+sin}A)\text{(cos}A)} \end{align}

\begin{align} \text{=}\frac{\text{2(1+sin}A)}{\text{(1+sin}A)\text{(cos}A)}\,\,\,\,\,\,\,\,\,\, \end{align} (By taking $$2$$ common and simplifying)

\begin{align} \text{=}\frac{\text{2}}{\text{cos}A} \end{align}

\begin{align} \text{=2sec}A \end{align}

=R.H.S.

(iii) \begin{align} \begin{bmatrix} \frac{\text{tan} \theta }{1-\text{cot} \theta} +\frac{\text{cot} \theta}{1-\text{tan} \theta}\\ =1+\text{sec} \theta \text{cosec} \theta \end{bmatrix} \end{align}

LHS\begin{align} {=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}} \dots(1) \end{align}

We know that the trigonometric functions,

\begin{align} & \tan \,(x)=\frac{\sin (x)}{\cos (x)} \\ & \cot \,(x)=\frac{\cos (x)}{\sin (x)}=\frac{1}{\tan (x)} \end{align}

By substituting the above relations in Equation $$(1)$$,

\begin{align} &=\frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\cos \theta }{\sin \theta }} \\ \\&=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin \theta -\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta -\sin \theta }{\cos \theta }} \\ \\ & = \begin{bmatrix} \frac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )} \\ + \frac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )} \end{bmatrix}\end{align}

Taking \begin{align} \,\,\frac{\text{1}}{\text{(sin }\!\!\theta\!\!\text{ }-\text{cos }\!\!\theta\!\!\text{ )}} \end{align} as common

\begin{align} \text{=}\,\,\frac{\text{1}}{\text{(sin }\!\!\theta\!\!\text{ }-\text{cos }\!\!\theta\!\!\text{ )}}\left[ \frac{\text{si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{cos }\!\!\theta\!\!\text{ }}-\frac{\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{sin }\!\!\theta\!\!\text{ }} \right] \\ \text{=}\,\,\frac{\text{1}}{\text{(sin }\!\!\theta\!\!\text{ }-\text{cos }\!\!\theta\!\!\text{ )}}\left[ \frac{\text{si}{{\text{n}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }-\text{co}{{\text{s}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }}{\text{sin}\,\text{ }\!\!\theta\!\!\text{ }\,\text{cos }\!\!\theta\!\!\text{ }} \right] \end{align}

Using

$$a^3 -b^3 =(a-b) \left( a^2 + \, ab \,+ \, b^2 \right)$$

\begin{align} =\frac{1}{\text{(sin}\theta-\text{cos} \theta)} \end{align} \begin{align} \begin{bmatrix} \frac{ \begin{pmatrix} \text{(sin} \theta -\text{cos} \theta) \\ \begin{pmatrix} \text{si}{{\text{n}}^{2}} \theta + \text{co}{{\text{s}}^2} \theta \\+\text{sin} \theta \text{cos } \theta \end{pmatrix} \end{pmatrix} }{\text{sin } \theta \text{cos } \theta } \end{bmatrix} \end{align}

\begin{align}\text{=}\frac{\text{(1}\,\text{+}\,\text{sin}\,\text{ }\!\!\theta\!\!\text{ }\,\text{cos}\,\text{ }\!\!\theta\!\!\text{ )}}{\text{(sin}\,\text{ }\!\!\theta\!\!\text{ cos}\,\text{ }\!\!\theta\!\!\text{ )}}\quad\end{align} (By identity $$\sin^2 A$$ $$+$$ $$\cos^2 A$$ $$= 1$$ )

\begin{align} \text{=1+}\,\text{sec}\,\text{ }\!\!\theta\!\!\text{ }\,\text{cosec}\,\text{ }\!\!\theta\!\!\text{ } \\ \end{align}

=R.H.S.

(iv) \begin{align} \frac{1+\sec A}{\sec A}&=\frac{\sin ^{2} A}{1-\cos A} \\ \end{align}

LHS \begin{align}=\frac{1+\sec A}{\sec A} \dots \dots (1) \end{align}

We know that the trigonometric functions,

\begin{align}\sec \,(x)=\frac{1}{\cos \,(x)}\end{align}

By substituting the above function in Equation $$(1)$$,

\begin{align} \frac{{{\rm{1}}\,{\rm{ + }}\,{\rm{sec}}\,{\rm{A}}}}{{{\rm{sec}}\,{\rm{A}}}}{\rm{ = }}\,\frac{{{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{cosA}}}}}}{{\frac{{\rm{1}}}{{{\rm{cosA}}}}}}\\ {\rm{ = }}\,\frac{{\frac{{{\rm{cosA + 1}}}}{{{\rm{cosA}}}}}}{{\frac{{\rm{1}}}{{{\rm{cosA}}}}}}\\ = \frac{{{\rm{cosA + 1}}}}{{{\rm{cosA}}}} \times \frac{{{\rm{cosA}}}}{1}\\ {\rm{ = }}\,{\rm{(1 + cosA)}} \end{align}

By multiplying $$\text{( 1 - cos A)}$$ in both denominator and numerator

\begin{align} &\Rightarrow\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)} \\ &=\frac{1-\cos ^{2} A}{1-\cos A} \\ &=\frac{\sin ^{2} A}{1-\cos A} \\ & \quad \begin{bmatrix} \text { By Identity } \\ \sin ^{2} A+\cos ^{2} A =1\end{bmatrix} \\ & = \text{R.H.S.} \end{align}

(v)\begin{align}\begin{bmatrix}\frac{\text{cos}\,\text{A}-\text{sin}\,\text{A+1}}{\text{cos}\,\text{A+sin}\,\text{A+1}}\\ =\text{cosec}\,\text{A}\,\text{+}\,\text{cotA} \end{bmatrix} \end{align}

$${\rm{L}}.{\rm{H}}.{\rm{S}} = \frac{{\cos {\rm{A}} - \sin {\rm{A}} + 1}}{{\cos {\rm{A}} + \sin {\rm{A}} - 1}}$$

Diving both numerator and denominator by $$\text{sin A}$$

\begin{align}\text{=}\frac{\frac{\text{cos}\,\text{A}}{\text{sin}\,\text{A}}-\frac{\text{sin}\,\text{A}}{\text{sin}\,\text{A}}\text{+}\frac{\text{1}}{\text{sinA}}}{\frac{\text{cos}\,\text{A}}{\text{sin}\,\text{A}}\text{+}\frac{\text{sin}\,\text{A}}{\text{sin}\,\text{A}}\text{+}\frac{\text{1}}{\text{sin}\,\text{A}}}\end{align}

We know that the trigonometric functions,

\,\begin{align} \text{cot}\,(x)= \frac{\text{cos}\,(x)}{\text{sin}\,(x)}\,\,\,&=\frac{1}{\text{tan}\,(x)} \\ \text{cosec}\,(x) &=\frac{\text{1}}{\text{sin}\,(x)} \end{align}

\,\begin{align} \text{We}\,\text{get} \\ \,\,\,&\Rightarrow =\frac{\text{cot}\,\text{A - 1+}\,\,\text{cosec}\,\text{A}}{\text{cot}\,\text{A}\,\text{+ 1- cosec}\,\text{A}} \\ &\Rightarrow =\frac{\text{cot}\,\text{A - (1-}\,\,\text{cosec}\,\text{A)}}{\text{cot}\,\text{A}\,\text{+ (1- cosec}\,\text{A)}}\end{align}

We know that,

\,\begin{align}\text{1 + co}{{\text{t}}^{\text{2}}}\text{A = Cose}{{\text{c}}^{\text{2}}}\text{A}\end{align}

Hence multiplying $$[\cot A – (1 – \rm{cosec}\; A)]$$ in numerator and denominator

\begin{align} & = \frac{ \begin{bmatrix} \{\text{ (cot}\,\text{A)}-(1-\text{cosec}\,\text{A) }\} \\ \{\text{ (cot}\,\text{A)}-\text{(1-cosec}\,\text{A) }\} \end{bmatrix} }{ \begin{bmatrix} \{\text{ (cot}\,\text{A)}\,+ \,(1-\text{cosec}\,\text{A) }\} \\ \{\text{ (cot}\,\text{A)}-(1-\text{cosec}\,\text{A) }\} \end{bmatrix} } \\ & = \frac{{{\text{(cot}\,\text{A}-1+ \text{cosec}\,\text{A)}}^{2}}}{{{\text{(cot}\,\text{A)}}^{2}}-{{(1-\text{cosecA)}}^{2}}} \\ & =\frac{ \begin{bmatrix} \text{co}{{\text{t}}^{2}}\text{A}\,+ 1+ \\ \text{cose}{{\text{c}}^{2}}\text{A}-2 \,\text{cot}\,\text{A} \\ -\text{2cosec}\,\text{A}\,\text{+} \\ \text{2}\,\text{cot}\,\text{A}\,\text{cosecA} \end{bmatrix} }{ \begin{bmatrix} \text{co}{{\text{t}}^{2}}\text{A} \\ -\begin{pmatrix} \text{1+cose}{{\text{c}}^{2}}\text{A} \\ -\text{2cosec}\,\text{A} \end{pmatrix} \end{bmatrix} } \\ & = \frac{ \begin{bmatrix}2\,\text{cose}{{\text{c}}^{2}}\text{A}\,+\,\text{2cot}\,\text{A}\,\,\text{cosecA} \\ - \text{2cot}\,\text{A}-\text{2cosecA} \end{bmatrix} }{ \begin{bmatrix} \text{co}{{\text{t}}^{2}}\text{A}-1+ \,\text{cose}{{\text{c}}^{\text{2}}}\text{A}\, \\ +\,\text{2}\,\text{cosecA} \end{bmatrix} }\,\, \\ & = \frac{ \begin{bmatrix} \text{2cosec}\,\text{A}\,\text{(cosec}\,\text{A}\,+ \,\text{cot}\,\text{A)}\,\\ -\text{2(cot}\,\text{A}-\text{cosec}\,\text{A)} \end{bmatrix} }{ \begin{bmatrix} \text{co}{{\text{t}}^{\text{2}}}\text{A}-\,\text{cose}{{\text{c}}^{2}}\text{A} \\ -1+2 \text{cosec}\,\text{A} \end{bmatrix} } \\ & = \frac{ \begin{bmatrix} \text{(cosec}\,\text{A} + \text{cotA)} \\ (2 \text{cosecA}-2) \end{bmatrix} }{ 1-1+ \,\text{2cosec}\,\text{A} } \\ & = \frac{ \begin{bmatrix} \text{(cosecA}\,+\,\text{cotA)} \\ (2\,\text{cosecA}\,-2) \end{bmatrix} }{(2 \,\text{cosec}\,\text{A}-2)} \\ & =\text{cosec A + cot A} \\ & =\text{R.H.S} \\ \end{align}

(vi) \begin{align} \sqrt{\frac{1+\sin A}{1-\sin A}}&=\sec A+\tan A \\\end{align}

LHS \begin{align} =\sqrt{\frac{1+\sin A}{1-\sin A}} \dots \dots (1) \end{align}

Multiplying and dividing by \begin{align}\sqrt{\left( \text{1+}\,\text{sin}\,\text{A} \right)}\end{align}

$\begin{array}{l} \Rightarrow \sqrt {\frac{{{\rm{(1 + }}\,{\rm{sin}}\,{\rm{A)(1 + }}\,{\rm{sin}}\,{\rm{A)}}}}{{{\rm{(1}} - {\rm{sin}}\,{\rm{A)(1 + }}\,{\rm{sin}}\,{\rm{A)}}}}} \\ = \sqrt {\frac{{{{{\rm{(1 + }}\,{\rm{sin}}\,{\rm{A)}}}^2}}}{{{\rm{(1}} - {\rm{si}}{{\rm{n}}^2}{\rm{A)}}}}} \\ \begin{bmatrix} a^2 - b^2 = \\ \left( {a-b} \right)\left( {a + b} \right), \end{bmatrix} \\ \,\begin{array}{*{20}{l}} \begin{array}{l} {\rm{ = }}\frac{{{\rm{(1 + }}\,{\rm{sinA)}}}}{{\sqrt {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A}}} }}\,\\ {\rm{ = }}\,\frac{{{\rm{1 + sin}}\,{\rm{A}}}}{{\sqrt {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{A}}} }} \end{array}\\ \begin{array}{l} {\rm{ = }}\frac{{{\rm{1 + }}\,{\rm{sin}}\,{\rm{A}}}}{{{\rm{cos}}\,{\rm{A}}}}\\ = \frac{1}{{{\rm{cos}}\,{\rm{A}}}}\,\, + \frac{{{\rm{sin}}\,{\rm{A}}}}{{{\rm{cos}}\,{\rm{A}}}}\\ {\rm{ = }}\,\,{\rm{sec}}\,{\rm{A}}\,{\rm{ + }}\,{\rm{tan}}\,{\rm{A}} \end{array}\\ {{\rm{ = R}}{\rm{.H}}{\rm{.S}}} \end{array} \end{array}$

(vii) \begin{align} { \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos \theta-\cos \theta}}&={\tan \theta} \end{align}

LHS \begin{align} &={\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}}\end{align}

Taking $$\text{Sin }\!\!\;\theta\!\!\;\text{ and Cos }\!\!\;\theta\!\!\text{ }$$ common in both numerator and denominator respectively.

\begin{align}=\Rightarrow \frac{\sin \text{ }\!\!\theta\!\!\text{ }\left( 1-2{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ } \right)}{\cos \text{ }\!\!\theta\!\!\text{ }\left( 2{{\cos }^{2}}\text{ }\!\!\theta\!\!\text{ }-1 \right)}\end{align}

By Identity $$\text{si}{{\text{n}}^{\text{2}}}\text{A + co}{{\text{s}}^{\text{2}}}\text{A = 1}$$ hence, $$\text{co}{{\text{s}}^{\text{2}}}\text{A = 1 - si}{{\text{n}}^{\text{2}}}\text{A}$$ and substituting this in the above equation,

$\begin{array}{l} \Rightarrow \frac{{{\rm{sin}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}{{{\rm{cos}}\,{\rm{\theta }}\left\{ {{\rm{2}}\left( {{\rm{1}} - {\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right) - {\rm{1}}} \right\}}}\\ = \frac{{{\rm{sin}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}{{{\rm{cos}}\,{\rm{\theta }}\left( {{\rm{2}} - 2{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }} - {\rm{1}}} \right)}}\\ {\rm{ = }}\,\frac{{{\rm{sin}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}{{{\rm{cos}}\,{\rm{\theta }}\left( {{\rm{1}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)}}\\ {\rm{ = }}\,\frac{{{\rm{sin}}\,{\rm{\theta }}}}{{{\rm{cos}}\,{\rm{\theta }}}}\,\\ {\rm{ = }}\,{\rm{tan}}\,{\rm{\theta }}\\ {\rm{ = RHS}} \end{array}$

(viii) \begin{align} \begin{bmatrix} {{\text{(sin} A\,\text{+}\,\text{cosec}\,A)}^{\text{2}}}\\+(cos\,A\,\text{+}\,\text{sec}\,A{{\text{)}}^{\text{2}}} \\ =\text{7+ta}{{\text{n}}^{\text{2}}}A\,\text{+}\,\text{co}{{\text{t}}^{\text{2}}}A \end{bmatrix} \end{align}

LHS \begin{align} = \begin{bmatrix} (\sin \mathrm{A}+\text{cosec} \mathrm{A})^{2}\\+(\cos \mathrm{A}+\sec \mathrm{A})^{2} \end{bmatrix} \end{align}

By using\begin{align}{{\left( \text{a + b } \right)}^{\text{2}}}&\text{ = }{{\text{a}}^{\text{2}}}\text{ + 2ab +}\,{{\text{b}}^{\text{2}}}\end{align}

\begin{align} \Rightarrow \quad & = \begin{bmatrix} \text{si}{{\text{n}}^{\text{2}}}\text{A}\,+ \,\text{cose}{{\text{c}}^{2}}\text{A}\,\\+ \,2 \,\text{sin}\,\text{A}\,\,\text{cosec}\,\text{A}\, \\ + \,\text{co}{{\text{s}}^{\text{2}}}\text{A}\,+ \,\text{se}{{\text{c}}^{2}}\text{A}\,\\+ \,2 \,\text{cosA}\,\,\,\text{secA} \end{bmatrix} \end{align}

By rearranging and using

\begin{align}\text{sec A } & = \frac{1}{\text{cos A}} \\ \\ & \qquad \rm and \\\\ \rm cosec A &= \frac{{1}}{\sin \rm A}\end{align}

\begin{align} \Rightarrow \begin{bmatrix} \left( \text{sin }^2\text{A}\,+ \,\text{co}{{\text{s}}^{2}}\text{A} \right) \\ + \left( \text{cose}{{\text{c}}^{2}}\text{A}\,+ \,\text{se}{{\text{c}}^{2}}\text{A} \right)\, \\ + \,2 \,\text{sin}\,\text{A}\left( \frac{1}{\text{sin}\,\text{A}} \right)\,\\ + \,\text{2cos}\,\text{A}\left( \frac{1}{\text{cos}\,\text{A}} \right) \end{bmatrix} \end{align}

Hence

\begin{align} & \left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A}}\,+ \,{\rm{co}}{{\rm{s}}^2}{\rm{A}}} \right)\,= \,1, \\ & {\rm{cose}}{{\rm{c}}^2}{\rm{A}}\, = \left( {{\rm{1 + co}}{{\rm{t}}^2}{\rm{A}}} \right)\\ & \qquad {\rm{ and }} \\ & \left( {{\rm{se}}{{\rm{c}}^2}{\rm{A}}\, - {\rm{ta}}{{\rm{n}}^2}{\rm{A}}} \right)\, = 1 \end{align}

\begin{align} & = \begin{bmatrix} (1)+\begin{pmatrix} 1+{{\cot }^{2}}\text{A}\\ +1+{{\tan }^{2}}\text{A} \end{pmatrix} \\ +(2)+(2) \end{bmatrix} \\ & =7+{{\tan }^{2}}\text{A}+{{\cot }^{2}}\text{A} \\ & = \text{R.H.S.} \end{align}

(ix) \begin{align} \begin{bmatrix} \text{(cosecA}-\text{sinA)}\\\text{(secA}-\text{cosA)} \\ =\frac{\text{1}}{\,\text{tan}\,\text{A+cot}\,\text{A}} \end{bmatrix} \end{align}

LHS \begin{align} = \begin{bmatrix} (\operatorname{cosec}\text{A}-\sin \text{A}) \\ (\sec \text{A}-\cos \text{A}) \end{bmatrix} \ldots (1) \\ \end{align}

We know that the trigonometric functions,

\begin{align} & \sec \,(x)=\frac{1}{\cos \,(x)} \\ & \operatorname{cosec}\,(x)=\frac{1}{\sin \,(x)} \end{align}

By substituting the above relations in Equation $$(1)$$

\begin{align} \Rightarrow \, & \begin{bmatrix} \left( {\frac{1}{{{\rm{sin}}\,{\rm{A}}}} - {\rm{sin}}\,{\rm{A}}} \right) \\ \left( {\frac{1}{{{\rm{cosA}}}} - {\rm{cos}}\,{\rm{A}}} \right) \end{bmatrix} \\ \\ &= \left( {\frac{{1 - {\rm{si}}{{\rm{n}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A}}}}} \right)\left( {\frac{{1 - {\rm{co}}{{\rm{s}}^2}{\rm{A}}}}{{{\rm{cos}}\,{\rm{A}}}}} \right)\\ & = \,\frac{{{\rm{co}}{{\rm{s}}^2}{\rm{A si}}{{\rm{n}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}\\ &= \frac{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}{1}\\ &= \frac{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A + co}}{{\rm{s}}^{\rm{2}}}{\rm{A}}}} \\ & \begin{bmatrix} {\left( {{\rm{si}}{{\rm{n}}^2 }{\rm{A}}\, + \,{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{A}}} \right)\, = 1} \end{bmatrix} \\ &= \frac{1}{{\frac{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{A + co}}{{\rm{s}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}}}\\ \\ & \begin{bmatrix} \text{Dividing numerator}\\ \text{ and denominator by } \\ \left( {\rm{sin}}\,{\rm{A cos}}\,{\rm{A}} \right) \end{bmatrix}\\ \\ &= \frac{1}{{\frac{{{\rm{si}}{{\rm{n}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}} + \frac{{{\rm{co}}{{\rm{s}}^2}{\rm{A}}}}{{{\rm{sin}}\,{\rm{A cos}}\,{\rm{A}}}}}}\\ &= \frac{1}{{\frac{{{\rm{sin A}}}}{{{\rm{cos}}\,{\rm{A}}}} + \frac{{{\rm{cos A}}}}{{{\rm{sin}}\,{\rm{A}}}}}}\\ &= \frac{1}{{\tan {\rm{A}} + \cot {\rm{A}}}}\\ &= {\rm{RHS}} \end{align}

(x) \begin{align}\,\,\left( \frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} \right)={{\left( \frac{1-\tan \text{A}}{1-\cot \text{A}} \right)}^{2}}\end{align}

Taking LHS, \begin{align}\left( \frac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)\end{align}

\begin{align} & \text{=}\,\frac{\text{se}{{\text{c}}^{\text{2}}}\text{A}}{\text{cose}{{\text{c}}^{\text{2}}}\text{A}} \\ & \text{=}\,\frac{\text{se}{{\text{c}}^{\text{2}}}\text{A}}{\text{cose}{{\text{c}}^{\text{2}}}\text{A}} \\ & \text{=}\,\frac{\frac{\text{1}}{\text{co}{{\text{s}}^{\text{2}}}\text{A}}}{\frac{\text{1}}{\text{si}{{\text{n}}^{\text{2}}}\text{A}}} \\ & \text{=}\,\frac{\text{1}}{\text{co}{{\text{s}}^{\text{2}}}\text{A}}\,\text{ }\!\!\times\!\!\text{ }\,\text{si}{{\text{n}}^{\text{2}}}\text{A} \\ & \text{=}\,\text{ta}{{\text{n}}^{\text{2}}}\text{A} \\&= \text{RHS} \\ \end{align}

Taking , \begin{align}{{\left( \frac{1-\tan \text{A}}{1-\cot \text{A}} \right)}^{2}}\end{align}

$$\begin{array}{c} {\rm{ = }}\,{\left( {\frac{{{\rm{1}} - {\rm{tanA}}}}{{{\rm{1}} - \frac{{\rm{1}}}{{{\rm{tanA}}}}}}} \right)^{\rm{2}}}\\ {\rm{ = }}\,{\left( {\frac{{{\rm{1}} - {\rm{tanA}}}}{{\frac{{{\rm{tanA}} - 1}}{{\tan {\rm{A}}}}}}} \right)^{\rm{2}}}\\ = {\left( {\left( {{\rm{1}} - {\rm{tanA}}} \right) \times \frac{{{\rm{tanA}}}}{{{\rm{tanA}} - 1}}} \right)^{\rm{2}}}\\ {\rm{ = }}\,{{\rm{(}} - {\rm{tanA)}}^{\rm{2}}}\\ {\rm{ = }}\,{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{A}}\\ {\rm{ = RHS}} \end{array}$$

Hence, L.H.S = R.H.S.

Ncert Class 10 Exercise 8.4
Ncert Solutions For Class 10 Maths Chapter 8 Exercise 8.4

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