Question

Prove that $\frac{{\csc }^{2}A+{\sec }^{2}A}{{\csc }^{2}A-{\sec }^{2}A}=\frac{1+{\tan }^{2}A}{1-{\tan }^{2}A}$

Answer 1

$\frac{\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A}}{\frac{1}{{\sin }^{2}A}-\frac{1}{{\cos }^{2}A}}=\frac{{\cos }^{2}A+{\sin }^{2}A}{{\cos }^{2}A-{\sin }^{2}A}=\frac{1}{{\cos }^{2}A-{\sin }^{2}A}=\frac{1}{\cos 2A}\to \left(1\right)$

$\frac{1+{\tan }^{2}A}{1-{\tan }^{2}A}=\frac{1+\frac{{\sin }^2}{{\cos }^2}}{1-\frac{{\sin }^2}{{\cos }^2}}=\frac{1}{{\cos }^{2}A-{\sin }^{2}A}=\frac{1}{\cos 2A}\to \left(2\right)$

(1) = (2) Hence Proved.