Question

Prove:

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

Answer 1

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

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

$={\tan }^2\theta$