How do you verify $\frac{1 - sin}{1 + sin} = {(sec - tan)}^{2}$ $(1-sin)/(1+sin) = (sec-tan)^2$ ?

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How do you verify $\frac{1 - sin}{1 + sin} = {(sec - tan)}^{2}$ $(1-sin)/(1+sin) = (sec-tan)^2$ ?

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Answer 1 · 2016-10-12T17:16:45

see below

Explanation:

$\frac{1 - sin x}{1 + sin x} = {(sec x - tan x)}^{2}$ $(1-sinx)/(1+sinx) = (secx-tanx)^2$

Left Side : $= \frac{1 - sin x}{1 + sin x}$ $=(1-sinx)/(1+sinx)$

$= \frac{1 - sin x}{1 + sin x} \cdot \frac{1 - sin x}{1 - sin x}$ $=(1-sinx)/(1+sinx) * (1-sinx)/(1-sinx)$

$= \frac{1 - 2 sin x + {sin}^{2} x}{1 - {sin}^{2} x}$ $=(1-2sinx+sin^2x)/(1-sin^2x)$

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

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

$= \frac{1}{{cos}^{2} x} - 2 \cdot \frac{1}{cos x} \frac{sin x}{cos x} + \frac{{sin}^{2} x}{{cos}^{2} x}$ $=1/cos^2x-2 * 1/cosx sinx/cosx+sin^2x/cos^2x$

$= {sec}^{2} x - 2 sec x tan x + {tan}^{2} x$ $=sec^2x-2secxtanx+tan^2x$

$= (sec x - tan x) (sec x - tan x)$ $=(secx-tanx)(secx-tanx)$

$= {(sec x - tan x)}^{2}$ $=(secx-tanx)^2$

$:.=$ Right Side

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Explanation:

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