Question

Prove that $\frac{{\tan }^{2}A}{{\tan }^{2}A+1}+\frac{cose{c}^{2}A}{{\sec }^{2}A+cose{c}^{2}A}=1$

Answer 1

$\frac{{\tan }^{2}A}{{\tan }^{2}A+1}+\frac{{\csc }^{2}A}{{\csc }^{2}A+{\sec }^{2}A}=1LHS=\frac{{\tan }^{2}A}{{\tan }^{2}A+1}+\frac{{\csc }^{2}A}{{\csc }^{2}A+{\sec }^{2}A}=\frac{\frac{{\sin }^{2}A}{{\cos }^{2}A}}{\frac{{\sin }^{2}A}{{\cos }^{2}A}+1}+\frac{\frac{1}{{\sin }^{2}A}}{\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A}}=\frac{{\sin }^{2}A}{{\sin }^{2}A+{\cos }^{2}A}+\frac{{\cos }^{2}A}{{\cos }^{2}A+{\sin }^{2}A}={\sin }^{2}A+{\cos }^{2}A=1=RHS$

LHS=RHS

Hence proof