Question

Simplify: $\frac{1-{\tan }^{2}x}{{\cot }^{2}x-1}$.

Answer 1

Now,

$\frac{1-{\tan }^{2}x}{{\cot }^{2}x-1}$

$=\frac{1-{\tan }^{2}x}{(\frac{1}{{\tan }^{2}x}-1)}$ [ Since, $\cot x=\frac{1}{\tan x}$ ]

$=\frac{{\tan }^{2}x(1-{\tan }^{2}x)}{(1-{\tan }^{2}x)}$

$={\tan }^{2}x$.