CBSE Class 10 Chapter 8 Introduction of Trigonometry Exercise 8.3

 

Q1.

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

Ans.

As we know that \(\cot A\) is reciprocal of \(\tan A\) So, \(\tan A = \frac {1}{\cot A}\) We also know that- \(1+\cot^{2}A = \text {cosec}^{2}A\) \(\text {cosec}^{2}A = 1+\cot^{2}A\) \(\text {cosec}A =\pm \sqrt {1+\cot^{2}A}\) Here, A is acute angle and \(\text {cosec}\) is positive for acute angle A. Therefore, \(\text {cosec}A =\sqrt {1+\cot^{2}A}........ (i)\) Expressing \(\bf\sin A\) We know that- \(\sin A = \frac {1}{\text {cosec }A}\) putting value from equation (i) \(\sin A = \frac {1}{\sqrt {1+\cot^{2}A}}\) Expressing \(\bf\sec A\) We know that- \(1+\tan^{2}A = \sec^{2}A\) \(\sec^{2}A=1+\tan^{2}A\) \(\sec A=\pm \sqrt {1+\tan^{2}A}\) Here, A is acute angle and \(\sec A \) is positive for acute angle A. Therefore, \(\sec A= \sqrt {1+\tan^{2}A}\) By converting \(\tan A\) to \(\cot A\) \(\sec A= \sqrt {1+\frac {1}{cot^{2}A}}\) \(\sec A= \sqrt {\left(\frac {\cot^{2}A+1}{\cot^{2}A}\right)}\) \(\sec A= \frac {\sqrt {\cot^{2}A+1}}{\cot A}\)

Q2.

Write all the other trigonometric ratios of \(\angle A\) in terms of \(\sec A\).

Ans.

\(\text {We know that}\) \(\cos A =\frac {1}{\sec A}\) And we also know that- \(1+\tan^{2}A = \sec^{2}A\) \(\text {By transposing}\) \(\tan^{2}A = \sec^{2}A-1\) \(\tan A = \pm\sqrt {\sec^{2}A-1}\) \(\text {Here, A is acute angle and A is positive for acute angle A}\). \(\tan A = \sqrt {\sec^{2}A-1}\) \(\text {We know that}\) \(\cot A = \frac {1}{\tan A}\) \(\text {Putting value}\) \(\cot A = \frac {1}{\sqrt{\sec^{2}A-1}}\) \(\text {We also know that}\) \(1+\cot^{2}A = \text {cosec}^{2}A\) \(\text {cosec}^{2}A=1+\cot^{2}A\) \(\text {Putting value}\) \(\text {cosec}^{2}A=1+\left(\frac {1}{\sqrt {\sec^{2}A-1}}\right)^{2}\) \(\text {cosec}^{2}A=1+\left(\frac {1}{\sec^{2}A-1}\right)\) \(\text {cosec}^{2}A=\left(\frac {1\times (\sec^{2}A-1)+1}{\sec^{2}A-1}\right)\) \(\text {cosec}^{2}A=\left(\frac {\sec^{2}A-1+1}{\sec^{2}A-1}\right)\) \(\text {cosec}^{2}A=\frac {\sec^{2}A}{\sec^{2}A-1}\) \(\text {cosec}A=\pm\sqrt{\frac{\sec^{2}A}{\sec^{2}A-1}}\) \(\text {cosec}A=\pm\frac{\sec^{2}A}{\sqrt{\sec^{2}A-1}}\) \(\text {Here, A is acute angle and \(\text {cosec}A\) is positive for acute angle A}\). \(\text {cosec}A=\frac{\sec^{2}A}{\sqrt{\sec^{2}A-1}}\) We know that- \(\sin A = \frac {1}{\text {cosec}A}\) \(\text {Putting value}\) \(\sin A = \frac{1}{\left(\frac {\sec A}{\sqrt {\sec^{2}A-1}}\right)}\) \(\sin A =\frac {\sqrt {\sec^{2}A-1}}{\sec A}\)

Q3.	Choose the correct option. Justify your choice.

(i)	\(\text {( i ) } 9\sec^{2}A-9\tan^{2}A =\) \(\text { (A)  }1 \text { (B)  } 9 \text { (C)  } 8 \text { (D)  } 0\)
Ans.	\begin{align*} \text {By using}\sec^{2}A = 1+\tan^{2}A\\ 9\sec^{2}A-9\tan^{2}A &=9(1+\tan^{2}A)-9\tan^{2}A\\ &=9+9\tan^{2}A-6\tan^{2}A\\ &=9+0\\ &=9\\ \text {Therefore, correct option is B} \end{align*}

(ii)

\((1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec}\theta)=\) \(\text { (A)  }0 \text { (B)  } 1 \text { (C)  } 2 \text { (D)  } -1\)

Ans.

\(\text {We know that-}\) \(\tan\theta=\frac {\sin\theta}{\cos\theta}\) \(\sec\theta=\frac {1}{\cos\theta}\) \(\cot\theta=\frac {\cos\theta}{\sin\theta}\) \(\text {cosec }\theta=\frac {1}{\sin\theta}\) \begin{align*} &\text {By putting values}\\ &=\left(1+\frac{\sin\theta}{\cos\theta}+\frac {1}{\cos\theta}\right)\left(1+\frac {\cos\theta}{\sin\theta}-\frac {1}{\sin\theta}\right)\\ &=\frac {\left(\cos\theta+\sin\theta+1\right)\left(1+\sin\theta+\cos\theta-1\right)}{\cos\theta\cdot\sin\theta}\\ &=\frac{\left(\cos\theta+\sin\theta\right)^{2}-1}{\cos\theta\cdot\sin\theta}\\ &=\frac{\cos^{2}\theta+\sin^{2}\theta+2\cos\theta\cdot\sin\theta-1}{\cos\theta\cdot\sin\theta}\\ &=\frac {1-1+2\cos\theta\cdot\sin\theta}{\cos\theta\cdot\sin\theta}\\ &=2 \end{align*} Therefore, correct option is C.

(iii)

\((\sec A+\tan A)(1-\sin A)=\) \(\text { (A)  }\sec A \text { (B)  } \sin A \text { (C)  } \text {cosec }A \text { (D)  } \cos A\)

Ans.

\(\text {We know that-}\) \(\tan A=\frac {\sin A}{\cos A}\) \(\sec A=\frac {1}{\cos A}\) \begin{align*} &=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\left(1-\sin A\right)\\ &=\left(\frac{1+\sin A}{\cos A}\right)\left(1-\sin A\right)\\ &=\frac {\left(1+\sin A\right)\left(1-\sin A\right)}{\cos A}\\ &\text {We know that-}(a+b)(a-b)=a^{2}-b^{2}\text { so}\\ &=\frac {\left(1^{2}-\sin^{2}A\right)}{\cos A}\\ &\text {We know that } \cos^{2}\theta = 1-\sin^{2}\theta\\ &=\frac {\cos^{2}A}{\cos A}\\ &=\cos A \end{align*} Therefore, correct option is D.

(iv)

\(\frac {1+\tan^{2}A}{1+\cot^{2}A}=\) \(\text { (A)  }\sec^{2} A \text { (B)  } -1 \text { (C)  } \cot^{2}A \text { (D)  } \tan^{2} A\)

Ans.

\(\text {We know that-}\) \(\tan A=\frac {\sin A}{\cos A}\) \(\cot A=\frac {\cos A}{\sin A}\) \begin{align*} \frac {1+\tan^{2}A}{1+\cot^{2}A} &=\frac{1+\frac{\sin^{2}A}{\cos^{2}A}}{1+\frac{\cos^{2}A}{\sin^{2}A}}\\ &=\frac {\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A}}{\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A}}\\ &=\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A}\times \frac{\cos^{2}A}{\cos^{2}A+\sin^{2}A}\\ & \text {We know that} \cos^{2}A+\sin^{2}A=1\\ &=\frac {1}{\cos^{2}A}\times \frac {\sin^{2}A}{1}\\ &= \frac {\sin^{2}A}{\cos^{2}A}\\ &=\left(\frac {\sin A}{\cos A}\right)^{2}\\ &=\bf\tan^{2}A \end{align*} Therefore, correct option is D.

Q4.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. \(\text {( i )} (\text {cosec}\theta - \cot\theta)^{2}=\frac {1-\cos\theta}{1+cos\theta}\) \(\text {( ii )} \frac {\cos A}{1+\sin A}+\frac {1+\sin A}{\cos A}=2\sec A\) \(\text {( iii )} \frac {\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\text {cosec}\theta\) \(\bf\text {Hint: Write the expression in terms of}\sin\theta\text{ and }\cos\theta\) \(\text {( iv )}\frac {1+\sec A}{\sec A}=\frac {\sin^{2}A}{1-\cos A}\) \(\bf\text {Hint : Simplify LHS and RHS separately}\) \(\text {( v )}\frac {\cos A-\sin A+1}{\cos A+\sin A-1}=\text {cosec}A + \cot A \text {, using the identity }\text {cosec}^{2}A=1+\cot^{2}A\). \(\text {( vi )}\sqrt{\frac {1+\sin A}{1-\sin A}}=\sec A + \tan A\) \(\text {( vii )}\frac {\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}=\tan\theta\) \(\text {( viii )}(\sin A+\text {cosec }A)^{2}+(\cos A+\sec A)^{2}=7+\tan^{2}A+\cot^{2}A\) \(\text {( ix )}(\text {cosec }A-\sin A)(\sec A-\cos A)=\frac {1}{\tan A+\cot A}\) \(\bf\text{Hint: Simplify LHS and RHS separately}\) \(\text {( x )}\left(\frac {1+\tan^{2}A}{1+\cot^{2}A}\right)=\left(\frac {1-\tan A}{1-\cot A}\right)^{2}=\tan^{2}A\)

(i)

\( (\text {cosec}\theta - \cot\theta)^{2}=\frac {1-\cos\theta}{1+cos\theta}\)

Ans.

\begin{align*} &\text {Write in terms of} \cos\theta \text { and } \sin\theta\\ (\text {cosec}\theta - \cot\theta)^{2}&=\left(\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\right)\\ &=\left(\frac {1-\cos\theta}{\sin\theta}\right)^{2}\\ &=\frac{\left(1-\cos\theta\right)^{2}}{\sin^{2}\theta}\\ &\text {We know that: }\sin^{2}\theta=1-\cos^{2}\theta\\ &=\frac{\left(1-\cos\theta\right)^{2}}{1-\cos^{2}\theta}\\ &=\frac{\left(1-\cos\theta\right)^{2}}{1^{2}-\cos^{2}\theta}\\ &\text {By using } a^{2}-b^{2}=(a-b)(a+b)\\ &=\frac {(1-\cos\theta)^{2}}{(1+\cos\theta)(1-\cos\theta)}\\ &=\frac {1-\cos\theta}{1+\cos\theta}\\ &=R.H.S. \end{align*}

(ii)

\( \frac {\cos A}{1+\sin A}+\frac {1+\sin A}{\cos A}=2\sec A\)

Ans.

\begin{align*} \frac {\cos A}{1+\sin A}+\frac {1+\sin A}{\cos A} &=\frac {\cos A(\cos A)+(1+\sin A)(1+\sin A)}{(1+\sin A)(\cos A)}\\ &=\frac{\cos^{2}A+(1+\sin A)^{2}}{(1+\sin A)(\cos A)}\\ &=\frac {\cos^{2}A+1^{2}+\sin^{2}A+2\sin A}{(1+\sin A)(\cos A)}\\ &=\frac {(\cos^{2}A+\sin^{2}A)+1+2\sin A}{(1+\sin A)(\cos A)}\\ &\text {We know that }\cos^{2}A+\sin^{2}A=1\\ &=\frac {1+1+2\sin A}{(1+\sin A)(\cos A)}\\ &=\frac {2+2\sin A}{(1+\sin A)(\cos A)}\\ &=\frac {2(1+\sin A)}{(1+\sin A)(\cos A)}\\ &=\frac {2}{\cos A}\\ &=2\times\frac {1}{\cos A}\\ &= 2\sec A\\ &=RHS \end{align*}

(iii)

\( \frac {\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\text {cosec}\theta\)

Ans.

\begin{align*} &\text {Covert LHS in terms of }\sin\theta \text { and } \cos\theta \\ \frac {\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}&=\frac {\frac {\sin\theta}{\cos\theta}}{1-\left(\frac{\cos\theta}{\sin\theta}\right)}+\frac {\left(\frac{\cos\theta}{\sin\theta}\right)}{1-\left(\frac{\sin\theta}{\cos\theta}\right)}\\ &=\frac {\frac{\sin\theta}{\cos\theta}}{\frac {\sin\theta-\cos\theta}{\sin\theta}}+\frac{\left(\frac {\cos\theta}{\sin\theta}\right)}{\left(\frac{\cos\theta-\sin\theta}{\cos\theta}\right)}\\ &=\frac {\sin\theta}{\cos\theta}\times\frac {\sin\theta}{\sin\theta-\cos\theta}+\frac {\cos\theta}{\sin\theta}\times\frac {\cos\theta}{\cos\theta-\sin\theta}\\ &=\frac {\sin^{2}\theta}{\cos\theta(\sin\theta-\cos\theta)}+\frac {\cos^{2}\theta}{\sin\theta(\cos\theta-\sin\theta)}\\ &=\frac {\sin^{2}\theta}{\cos\theta(\sin\theta-\cos\theta)}+\frac {\cos^{2}\theta}{\sin\theta\times - (\sin\theta-\cos\theta)}\\ &=\frac {\sin^{2}\theta}{\cos\theta(\sin\theta-\cos\theta)}-\frac {\cos^{2}\theta}{\sin\theta(\sin\theta-\cos\theta)}\\ &=\frac {\sin^{2}\theta\times (\sin\theta)-\cos^{2}\theta\times(\cos\theta)}{\cos\theta(\sin\theta-\cos\theta)\sin\theta}\\ &=\frac {\sin^{3}\theta-\cos^{3}\theta}{\cos\theta\sin\theta(\sin\theta-\cos\theta)}\\ & \text {By using } a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)\\ &=\frac {(\sin\theta-\cos\theta)(\sin^{2}\theta+\cos^{2}\theta+\cos\theta\sin\theta)}{\cos\theta\sin\theta(\sin\theta-\cos\theta)}\\ &=\frac {\sin^{2}\theta+\cos^{2}\theta+\cos\theta\sin\theta}{\cos\theta\sin\theta}\\ & \text {We know that } \cos^{2}\theta+\sin^{2}\theta = 1\\ &= \frac {1+\cos\theta\sin\theta}{\cos\theta\sin\theta}\\ &=\frac {1}{\cos\theta\sin\theta}+\frac {\cos\theta\sin\theta}{\cos\theta\sin\theta}\\ &=\frac {1}{\cos\theta}\times \frac {1}{\sin\theta}+1 \\ &= \sec\theta\times\text{cosec}\theta+1\\ &=1+\sec\theta\text {cosec}\theta\\ &= RHS \end{align*}

(iv)

\(\frac {1+\sec A}{\sec A}=\frac {\sin^{2}A}{1-\cos A}\)

Ans.

\begin{align*} \text {LHS}\\ \frac{1+\sec A}{\sec A} &= \frac {1}{\sec A}+\frac {\sec A}{\sec A}\\ &= \frac {1}{\sec A}+1\\ &\text {We know that} \cos A =\frac {1}{\sec A}\\ &= \cos A +1 \end{align*} \begin{align*} &\text {RHS}\\ &=\frac {\sin^{2}A}{1-\cos A}\\ &\text {We know that} \sin^{2}\theta = 1-\cos^{2}\theta \\ &= \frac {1-\cos^{2}A}{1-\cos A}\\ &= \frac {1^{2}-\cos^{2}A}{1-\cos A}\\ &\text {By using} a^{2}-b^{2}=(a-b)(a+b)\\ &= \frac {(1-\cos A)(1+\cos A)}{1-\cos A}\\ &=1+cos A \end{align*}

(v)

\(\frac {\cos A-\sin A+1}{\cos A+\sin A-1}=\text {cosec}A + \cot A \text {, using the identity }\text {cosec}^{2}A=1+\cot^{2}A\).

Ans.

\begin{align*} &\text {LHS}\\ &\text {By divideing numerator and denominator by } \sin A\\ \frac {\cos A-\sin A+1}{\cos A+\sin A-1}&=\frac{\frac{\cos A}{\sin A}-\frac {\sin A}{\sin A}+\frac {1}{\sin A}}{\frac {\cos A}{\sin A}+\frac {\sin A}{\sin A}-\frac {1}{\sin A}}\\ &=\frac {\cot A-1+\text {cosec }A}{\cot A+1-\text {cosec }A}\\ &=\frac {(\cot A + \text {cosec }A)-1}{(\cot A + 1-\text {cosec }A)}\\ &\text {We know that } \text {cosec}^{2}\theta - \cot^{2}\theta =1\\ &= \frac {(\cot A+\text {cosec }A)-(\text {cosec}^{2}A-\cot^{2}A)}{(\cot A+1-\text {cosec }A)}\\ &\text {By using} a^{2}-b^{2}=(a-b)(a+b)\\ &=\frac {(\cot A+\text {cosec }A)-(\text {cosec}A-\cot A)(\text {cosec }A+\cot A)}{\cot A+1-\text {cosec }A}\\ &= \frac {(\cot A+\text {cosec}A)[1-(\text {cosec}A-\cot A)]}{[\cot A+1-\text {cosec}A]}\\ &=\frac {(\cot A+\text {cosec}A)[1-\text {cosec}A+\cot A]}{[\cot A+1-\text {cosec}A]}\\ &=\frac {(\cot A+\text {cosec}A)[\cot A+1-\text {cosec}A]}{[\cot A+1-\text {cosec}A]}\\ &=\cot A+\text {cosec }A\\ &= RHS \end{align*}

(vi)

\(\sqrt{\frac {1+\sin A}{1-\sin A}}=\sec A + \tan A\)

Ans.

\begin{align*} &\text {LHS}\\ &\text {To rationalising denominator}\\ &\text {Multiply numerator and denominator by } (1+\sin A)\\ \sqrt{\frac {1+\sin A}{1-\sin A}}&=\sqrt{\frac {(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}\\ &=\sqrt {\frac {(1+\sin A)^{2}}{1^{2}-\sin^{2} A}}\\ &=\sqrt {\frac {(1+\sin A)^{2}}{1-\sin^{2} A}}\\ &\text {We know that } 1-\sin^{2}A=\cos^{2}A\\ &=\sqrt {\frac {(1+\sin A)^{2}}{\cos^{2} A}}\\ &=\sqrt {\left(\frac {1+\sin A}{\cos A}\right)^{2}}\\ &=\frac {1+\sin A}{\cos A}\\ &=\frac {1}{\cos A}+\frac {\sin A}{\cos A}\\ &= \sec A + tan A\\ &= RHS \end{align*}

(vii)

\(\frac {\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}=\tan\theta\)

Ans.

\begin{align*} &\text {LHS}\\ \frac {\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}&=\frac {\sin\theta(1-2\sin^{2}\theta)}{\cos\theta(2\cos^{2}\theta-1)}\\ &=\frac {\sin\theta}{\cos\theta}\times \frac {(1-2\sin^{2}\theta)}{(2\cos^{2}\theta-1)}\\ &\text {We know that} \tan\theta = \frac {\sin\theta}{\cos\theta}\\ &= \tan\theta\times \frac {(1-2\sin^{2}\theta)}{(2\cos^{2}\theta-1)}\\ &\text {We know that} \cos^{2}\theta = 1-\sin^{2}\theta\\ &=\tan\theta\times \frac {(1-2\sin^{2}\theta)}{[2(1-2\sin^{2}\theta)-1]}\\ &=\tan\theta\times \frac {(1-2\sin^{2}\theta)}{(2-2\sin^{2}\theta-1}\\ &=\tan\theta\times \frac {(1-2sin^{2}\theta)}{(1-2sin^{2}\theta)}\\ &=\tan\theta\times1\\ &= tan\theta\\ &=RHS \end{align*}

(viii)

\((\sin A+\text {cosec }A)^{2}+(\cos A+\sec A)^{2}=7+\tan^{2}A+\cot^{2}A\)

Ans.

\begin{align*} &\text {LHS}\\ &\sin A+\text {cosec }A)^{2}+(\cos A+\sec A)^{2}\\ &=\left(\sin^{2}A+\text{cosec}^{2}A+2\sin A\cdot\text{ cosec }A\right)+\left(\cos^{2}A+\sec^{2}A+2\cos A\cdot\sec A\right)\\ &=\left(\sin^{2}A+\text{cosec}^{2}A+2\sin A\cdot\frac {1}{\sin A}\right)+\left(\cos^{2}A+\sec^{2}A+2\cos A\cdot\frac {1}{\cos A}\right)\\ &= \left(\sin^{2}A+\text{cosec}^{2}A+2\right)+\left(\cos^{2}A+\sec^{2}A+2\right)\\ & \text {We know that }\text {cosec}^{2}A=1+\cot^{2}A \text { and } \sec^{2}A = 1+\tan^{2}A\\ &=\left(\sin^{2}A+(1+\cot^{2}A)+2\right)+\left(\cos^{2}A+(1+\tan^{2}A)+2\right)\\ &=\sin^{2}A+\cot^{2}A+1+2+\cos^{2}A+1+\tan^{2}A+2\\ &=(\sin^{2}A++\cos^{2}A)+\cot^{2}A+\tan^{2}A+(1+2+1+2)\\ & \text {By using} \sin^{2}A+\cos^{2}A=1\\ &=1+\cot^{2}A+\tan^{2}A+6\\ &=7+\cot^{2}A+\tan^{2}A\\ &=RHS \end{align*}

(ix)

\((\text {cosec }A-\sin A)(\sec A-\cos A)=\frac {1}{\tan A+\cot A}\)

Ans.

\(\text {LHS}\) \( (\text {cosec }A-\sin A)(\sec A-\cos A)\) \(=\left(\frac {1}{\sin A}-\sin A\right)\left(\frac {1}{\cos A}-\cos A\right)\) \(=\frac {(1-\sin^{2}A)}{\sin A}\times \frac {(1-\cos^{2}A)}{\cos A}\) \(\text {We know that} \cos^{2}\theta = 1-\sin^{2}\theta \text { and }\sin^{2}\theta = 1-\cos^{2}\theta\) \(\frac {\cos^{2}A}{\sin A}\times \frac {\sin^{2}A}{\cos^{2}A}\) \(\sin A\cdot\cos A\) \( \text {RHS}\) \( \frac {1}{\tan A+\cot A}\) \(=\frac {1}{\frac {\sin A}{\cos A}+\frac {\cos A}{\sin A}}\) \(=\frac {1}{\frac {\sin A(\sin A)+\cos A (\cos A)}{\cos A\sin A}}\) \(=\frac {1}{\frac {\sin^{2}A+\cos^{2}A}{\cos A\cdot\sin A}}\) \(=\frac {\sin A\cdot\cos A}{\sin^{2}A+\cos^{2}A}\) \( \text {By using } \sin^{2}A+\cos^{2}A=1\) \(=\frac {\sin A\cdot\cos A}{1}\) \(= \sin A\cdot\cos A\) \(= LHS\)

(x)

\(\left(\frac {1+\tan^{2}A}{1+\cot^{2}A}\right)=\left(\frac {1-\tan A}{1-\cot A}\right)^{2}=\tan^{2}A\)

Ans.

\begin{align*} &\text {By solving}\\ &=\left(\frac {1+\tan^{2}A}{1+\cot^{2}A}\right)\\ &=\frac {(1+\tan^{2}A)}{\left(1+\frac {1}{\tan^{2}A}\right)}\\ &=\frac {(1+\tan^{2}A)}{\frac {(\tan^{2}A+1)}{\tan^{2}A}}\\ &=\frac {\tan^{2}A(1+\tan^{2}A)}{(\tan^{2}A+1)}\\ &=\bf\tan^{2}A\\ &=RHS &\text {By solving}\\ &=\left(\frac {1-\tan A}{1-\cot A}\right)^{2}\\ &=\left(\frac {1-\tan A}{1-\frac {1}{\tan A}}\right)^{2}\\ &=\left(\frac {1-\tan A}{\frac {(\tan A-1)}{\tan A}}\right)^{2}\\ &=\left(\frac {\tan A(1-\tan A)}{(\tan A-1)}\right)^{2}\\ &=\left(\frac {\tan A(1-\tan A)}{-(1-\tan A)}\right)^{2}\\ &=\left(-\tan A\right)^{2}\\ &=\tan^{2}A\\ &=RHS\\ &=\text {Hence Proved} \end{align*}