Question

Prove that:

$\frac{1+ta{n}^2\Theta }{1+co{t}^2\Theta }=(\frac{1-tan\Theta }{1-cot\Theta }{)}^2=ta{n}^2\Theta .$

Answer 1

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

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

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

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

$=(-\tan \theta {)}^2$

$={\tan }^2\theta$.