Question

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

Answer 1

To show: $\left(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}\right)={\left(\frac{1-\tan A}{1-\cot A}\right)}^2={\tan }^{\prime }A$

Considers $\left(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}\right)$

$=\frac{se{c}^{2}A}{cose{c}^{2}A}$ $\left[\begin{array}{c}{\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}\right]$

$=\frac{{\sin }^{2}A}{{\cos }^2}={\tan }^{2}A$

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

Consider ${\left(\frac{1-\tan A}{1-\cot A}\right)}^2$

$={\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)}^2[\cot \theta =\frac{1}{\tan \theta }]$

${\left(\frac{1-\tan A}{(\tan A-1).\frac{1}{\tan A}}\right)}^2$

${\left(\frac{-(\tan A-1).\tan A}{(\tan A-1)}\right)}^2$

$ta{n}^{2}A$

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

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