Question

The expression $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as :

Answer 1

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

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

$=$$\frac{tanA}{\frac{tanA-1}{tanA}}+\frac{1}{tanA(1-tanA)}$

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

$=$$\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}$

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

$=$$tanA+1+cotA$

So,$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as $tanA+1+cotA$