# How do you prove (tan x / (1 - cot x)) + (cot x / (1 - tan x)) = 1 + sec x csc x?

Jun 2, 2018

$L H S = \left(\tan \frac{x}{1 - \cot x}\right) + \left(\cot \frac{x}{1 - \tan x}\right)$

$= \left(\tan \frac{x}{1 - \cot x}\right) + \left(\cot \frac{x}{1 - \frac{1}{\cot} x}\right)$

$= \left(\frac{\frac{1}{\cot} x}{1 - \cot x}\right) - \left({\cot}^{2} \frac{x}{1 - \cot x}\right)$

$= \frac{\frac{1}{\cot} x - {\cot}^{2} x}{1 - \cot x}$

$= \frac{1 - {\cot}^{3} x}{\cot x \left(1 - \cot x\right)}$

$= \frac{\left(1 - \cot x\right) \left(1 + \cot x + {\cot}^{2} x\right)}{\cot x \left(1 - \cot x\right)}$

$= \frac{\cancel{\left(1 - \cot x\right)} \left(1 + \cot x + {\cot}^{2} x\right)}{\cot x \cancel{\left(1 - \cot x\right)}}$

$= \frac{1 + \cot x + {\cot}^{2} x}{\cot} x$

$= \frac{\cot x + {\csc}^{2} x}{\cot} x$

$= \cot \frac{x}{\cot} x + {\csc}^{2} \frac{x}{\cot} x$

=1+(1/sin^2x) / (cosx/sinx_

$= 1 + \frac{1}{\sin} ^ 2 x \times \sin \frac{x}{\cos} x$

$= 1 + \frac{1}{\sin x \cos x}$

$= 1 + \sec x \csc x = R H S$