Question

Prove that $\frac{{\tan }^{2}A}{1+{\tan }^{2}A}+\frac{{\cot }^{2}A}{1+{\cot }^{2}A}=1$

Answer 1

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

Using the identites: $1+{\tan }^{2}A={\sec }^{2}A$ and $1+{\cot }^{2}A={\csc }^{2}A$:

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

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

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

$=1$

$=RHS$

Hence proved.