Question

Prove:

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

Answer 1

Lets evaluate, $\left[\frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }\right]$

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

$=\frac{(1+{\tan }^2\theta ){\tan }^2\theta }{({\tan }^2\theta +1)})$

$={\tan }^2\theta$

$\Rightarrow \frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }={\tan }^2\theta$ ---(1)

Now, lets evaluate, ${\left[\frac{1-\tan \theta }{1-\cot \theta }\right]}^2$

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

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

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

$=[-\tan \theta {]}^2$

$={\tan }^2\theta$

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

From (1) and (2),

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

Hence, proved.