Question #029c3

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Question #029c3

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Answer 1 · 2017-10-02T06:35:10

Proved $\frac{1}{cos x (1 + cos x)} = \frac{tan x - sin x}{{sin}^{3} x}$ $1/[cos x(1+cos x)]= [tan x -sin x]/sin^3x$

Explanation:

Given, $\frac{1}{cos x (1 + cos x)} = \frac{tan x - sin x}{{sin}^{3} x}$ $1/[cos x(1+cos x)] = [tan x - sin x]/sin^3x$

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

$\frac{1}{cos x (1 + cos x)}$ $1/[cos x (1+cos x)]$

$\Rightarrow \frac{1 . (1 - cos x)}{cos x (1 + cos x) (1 - cos x)}$ $rArr [1.(1-cos x)]/[cos x (1+cos x)(1-cos x)$ [ multiply both sides by (1 - cos x)]

$\Rightarrow \frac{1 - cos x}{cos x (1 - {cos}^{2} x)}$ $rArr [1 - cos x]/[cos x (1 - cos^2 x)]$

$\Rightarrow \frac{\frac{1 - cos x}{cos x}}{{sin}^{2} x}$ $rArr [(1-cos x)/cos x]/sin^2x$ [ as ${sin}^{2} x + {cos}^{2} x = 1 . s o, 1 - {cos}^{2} x = {sin}^{2} x]$ $sin^2x + cos^2x = 1. so, 1-cos^2x = sin^2x]$

$\Rightarrow \frac{\frac{1}{cos x} - \frac{cos x}{cos x}}{{sin}^{2} x}$ $rArr [1/cos x - cos x/cos x]/sin^2x$

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

$\Rightarrow \frac{\frac{sin x}{cos x} - sin x}{{sin}^{3} x}$ $rArr [sin x/cos x - sin x]/sin^3x$

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

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Question #029c3

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