Question

Prove that:
$\frac{\tan A}{(1+{\tan }^{2}A{)}^2}+\frac{\cot A}{(1+{\cot }^{2}A{)}^2}=\sin A\cos A$

Answer 1

LHS$=\frac{\tan A}{(1+{\tan }^{2}A{)}^2}+\frac{\cot A}{(1+{\cot }^{2}A{)}^2}$

$\frac{\tan A}{({\sec }^{2}A{)}^2}+\frac{\cot A}{(cose{c}^{2}A{)}^2}$

$=\frac{\frac{\sin A}{\cos A}}{\frac{1}{{\cos }^{4}A}}+\frac{\frac{\cos A}{\sin A}}{\frac{1}{{\sin }^{4}A}}$

$=\sin A{\cos }^{3}A+\cos A{\sin }^{3}A$

$=\sin A\cos A[{\sin }^{2}A+{\cos }^{2}A]$

$=\sin A\cos A=RHS$

Hence proved.