Question

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

Answer 1

Simplifying the LHS of $\frac{1-{\tan }^{2}A}{{\cot }^{2}A-1}={\tan }^{2}A$.

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

$=\frac{\frac{{\cos }^{2}A-{\sin }^{2}A}{{\cos }^{2}A}}{\frac{{\cos }^{2}A-{\sin }^{2}A}{{\sin }^{2}A}}$

$=\frac{{\sin }^{2}A}{{\cos }^{2}A}$

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

$=RHS$