# Question 029c3

Oct 2, 2017

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

#### Explanation:

Given, $\frac{1}{\cos x \left(1 + \cos x\right)} = \frac{\tan x - \sin x}{\sin} ^ 3 x$

We have to prove either way. Let me take Left Hand Side (L.H.S.)

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

rArr [1.(1-cos x)]/[cos x (1+cos x)(1-cos x) [ multiply both sides by (1 - cos x)]

$\Rightarrow \frac{1 - \cos x}{\cos x \left(1 - {\cos}^{2} x\right)}$

$\Rightarrow \frac{\frac{1 - \cos x}{\cos} x}{\sin} ^ 2 x$ [ as sin^2x + cos^2x = 1. so, 1-cos^2x = sin^2x]#

$\Rightarrow \frac{\frac{1}{\cos} x - \cos \frac{x}{\cos} x}{\sin} ^ 2 x$

$\Rightarrow \frac{\left(\frac{1}{\cos} x - 1\right) \sin x}{{\sin}^{2} x . \sin x}$ [ multiply both sides by sin x]

$\Rightarrow \frac{\sin \frac{x}{\cos} x - \sin x}{\sin} ^ 3 x$

$\Rightarrow \frac{\tan x - \sin x}{\sin} ^ 3 x$ = R. H. S.