Question

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

Answer 1

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

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

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

LHS=RHS

Hence proved.