Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{1+tan\alpha +ta{n}^2\alpha }{1+cot\alpha +co{t}^2\alpha }=ta{n}^2\alpha .$

Answer 1

L.H.S.

$\frac{1+\tan \alpha +{\tan }^2\alpha }{1+\cot \alpha +{\cot }^2\alpha }$

$=\frac{\tan \alpha +{\sec }^2\alpha }{\cot \alpha +{\csc }^2\alpha }$

$=\frac{\frac{\sin \alpha }{\cos \alpha }+\frac{1}{{\cos }^2\alpha }}{\frac{\cos \alpha }{\sin \alpha }+\frac{1}{{\sin }^2\alpha }}$

$\frac{\frac{\sin \alpha \cos \alpha +1}{{\cos }^2\alpha }}{\frac{\sin \alpha \cos \alpha +1}{{\sin }^2\alpha }}$

$=\frac{\sin \alpha \cos \alpha +1}{{\cos }^2\alpha }\times \frac{{\sin }^2\alpha }{\sin \alpha \cos \alpha +1}$

$=\frac{{\sin }^2\alpha }{{\cos }^2\alpha }$

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