Question

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

Answer 1

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

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

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

$={\tan }^{2}A=$ R.H.S

Hence, proved