Question

Prove that $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\tan A+\cot A$.

Answer 1

$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$

As $cotA=\frac{1}{tanA}$

$\frac{ta{n}^{2}A}{tanA-1}-\frac{1}{tanA(tanA-1)}$

$\frac{ta{n}^{3}A-1}{tanA(tanA-1)}$

$\frac{(tanA-1)(ta{n}^{2}A+tanA+1)}{tanA(tanA-1)}$

$\frac{ta{n}^{2}A+tanA+1}{tanA}$

$=tanA+1+cotA$

$=1+tanA+cotA$

$\therefore$ Proved.