# Tanx/1-cotx+cotx/1-tanx=?

Mar 29, 2018

$\tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x}$

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

$= \frac{- {\tan}^{2} x}{1 - \tan x} + \cot \frac{x}{1 - \tan x}$ [because" tanx xx cotx = 1]

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

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

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

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

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

=1/tanx+tanx/tanx+tan^2x/tanx)

=color(red)(1+tanx+cotx

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

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

=color(red)((sinxcosx + 1)/ (sinxcosx)

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

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

=color(red)( 1 + 2/(sin2x)

The final answer may vary, depending on where you may wanna stop :)

Mar 29, 2018

$\tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x} = \sec x \csc x + 1$.

#### Explanation:

$\tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x}$,

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

$= \left\{\sin \frac{x}{\cos} x \div \frac{\sin x - \cos x}{\sin} x\right\} + \left\{\cos \frac{x}{\sin} x \div \frac{\cos x - \sin x}{\cos} x\right\}$,

$= \left\{\sin \frac{x}{\cos} x \times \sin \frac{x}{\sin x - \cos x}\right\} + \left\{\cos \frac{x}{\sin} x \times \cos \frac{x}{\cos x - \sin x}\right\}$,

$= {\sin}^{2} \frac{x}{\cos x \left(\sin x - \cos x\right)} - {\cos}^{2} \frac{x}{\sin x \left(\sin x - \cos x\right)}$,

$= \frac{1}{\sin x - \cos x} \left\{{\sin}^{2} \frac{x}{\cos} x - {\cos}^{2} \frac{x}{\sin} x\right\}$,

$= \frac{1}{\sin x - \cos x} \left\{\frac{{\sin}^{3} x - {\cos}^{3} x}{\sin x \cos x}\right\}$,

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

$= \frac{\left({\sin}^{2} x + {\cos}^{2} x\right) + \sin x \cos x}{\sin x \cos x}$,

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

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

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

$\Rightarrow \tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x} = \sec x \csc x + 1$.