Question

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

Answer 1

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

$=\frac{1-\frac{1}{{\cot }^{2}A}}{{\cot }^{2}A-1}$

$=\frac{{\cot }^{2}A-1}{{\cot }^{2}A({\cot }^{2}A-1)}$

$={\tan }^{2}A$

$=RHS$

$\because L.H.S=R.H.S$

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

Hence proved