Question

Prove that:
$\frac{\tan A}{1-\cot A}+\frac{\cos A}{1-\tan A}=1+\tan A+\cot A=1+\sec A\csc A$

Answer 1

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

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

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

$=\frac{{\tan }^{2}A-1/tanA}{(tanA-1)}$

$=\frac{{\tan }^{3}A-1}{tanA(tanA-1)}$

$=\frac{(tanA-1)({\tan }^{2}A+1+tanA)}{tanA(tanA-1)}$

$=\frac{{\tan }^{2}A+1+tanA}{tanA}=tanA+cotA+1$

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

Now,

$tanA+cotA+1=\frac{sinA}{cosA}+\frac{cosA}{sinA}+1=\frac{{\sin }^{2}A+{\cos }^{2}A}{sinA.cosA}+1$

$=\frac{1}{sinA.cosA}+1$

$=cosecA.secA+1$

$\therefore$ $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=tanA+cotA+1=cosecA.secA+1$