Question #0038f

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Question #0038f

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Answer 1 · 2017-02-18T08:26:59

see explanation

Explanation:

$\frac{1}{sec x - tan x} + \frac{1}{sec x + tan x} = \frac{2}{cos x}$ $1/(sec x - tanx) + 1/(secx+tanx) = 2/cosx$

equate the denominator

$\frac{sec x + tan x + sec x - tan x}{(sec x - tan x) \cdot (sec x + tan x)} = \frac{2}{cos x}$ $(sec x+tan x + sec x - tan x)/((sec x - tanx) *(secx+tanx)) = 2/cosx$

$\frac{2 sec x}{{sec}^{2} x - {tan}^{2} x} = \frac{2}{cos x}$ $(2sec x) /(sec ^2x - tan ^2x) = 2/cos x$

make sec and tan become fraction

$\frac{2 sec x}{\frac{1}{{cos}^{2} x} - \frac{{sin}^{2} x}{{cos}^{2} x}} = \frac{2}{cos x}$ $(2sec x )/ (1/cos^2 x - sin^2 x/cos ^2 x)= 2/cosx$

$\frac{2 sec x}{\frac{1 - {sin}^{2} x}{{cos}^{2} x}} = \frac{2}{cos x}$ $(2sec x )/((1 - sin^2 x)/cos ^2 x)= 2/cosx$

use trig identity so $1 - {sin}^{2} x = {cos}^{2} x$ $1-sin^2x =cos ^ 2 x$

$\frac{2 sec x}{\frac{{cos}^{2} x}{{cos}^{2} x}} = \frac{2}{cos x}$ $(2sec x )/((cos^2x)/cos ^2 x)= 2/cosx$

$\frac{2 sec x}{1} = \frac{2}{cos x}$ $(2sec x )/ 1= 2/cosx$

$2 \cdot \frac{1}{cos x} = \frac{2}{cos x}$ $2 * 1/cos x = 2/cos x$

$\frac{2}{cos x} = \frac{2}{cos x}$ $2/cosx = 2/cos x$ (verified)

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Question #0038f

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