Question

Demonstrate that $\frac{1}{{\sin }^{2}x}$ is equivalent to $\frac{(1+{\tan }^{2}x)}{{\tan }^{2}x}$

Answer 1

The equation $\frac{1+{\tan }^{2}x}{{\tan }^{2}x}$ is equivalent to

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

$=\frac{{\cos }^{2}x+{\sin }^{2}x}{{\sin }^{2}x}$

$=\frac{1}{{\sin }^{2}x}$