# How do you verify tanx / (1-cosx) = cscx (1+secx)?

Apr 22, 2016

Use trick algebra while working with the RHS.

#### Explanation:

There probably is a more efficient method. Nonetheless, here is my work to make the right hand side look like the left:

RHS $=$$\csc x \left(1 + \sec x\right)$

$=$ $\csc x + \csc x \sec x$ --distribution
$=$ $\frac{1}{\sin} x + \frac{1}{\cos x \sin x}$ -- definition

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

Multiply both numerator and denominator by (cos x -1)
$\Rightarrow$ $\frac{- \left(1 - {\cos}^{2} x\right)}{{\cos}^{2} x \sin x - \cos x \sin x}$

--also took out a negative to fit with Pythagorean identity 1

Then multiply both the numerator and the denominator by (1/sin x )
$\Rightarrow$$\frac{- \sin x}{{\cos}^{2} x - \cos x}$

$=$$\frac{- \sin x}{\cos x \left(\cos x - 1\right)}$

Now do same with (1/cos x )
$\Rightarrow$$\frac{\tan x}{-} \left(\cos x - 1\right)$ --moved negative to bottom

$=$$\frac{\tan x}{1 - \cos x}$$=$ LHS, that we needed to verify!

Hope this helps :-)

May 20, 2017

$L H S = \tan \frac{x}{1 - \cos x}$

=(tanx(1+cosx))/((1-cosx)(1+cosx)#

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

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

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

$= \left(\frac{1}{\sin} x\right) \times \left(\frac{1}{\cos} x + \cos \frac{x}{\cos} x\right)$

$= \csc x \times \left(\sec x + 1\right) = R H S$

Proved