How do you prove $4 tan x sec x = (\frac{1 + sin x}{1 - sin x}) - (\frac{1 - sin x}{1 + sin x})$ $4tanxsecx=((1+sinx)/(1-sinx))-((1-sinx)/(1+sinx))$ ?

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How do you prove $4 tan x sec x = (\frac{1 + sin x}{1 - sin x}) - (\frac{1 - sin x}{1 + sin x})$ $4tanxsecx=((1+sinx)/(1-sinx))-((1-sinx)/(1+sinx))$ ?

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Answer 1 · 2016-11-05T16:34:25

$R H S = \frac{1 + sin x}{1 - sin x} - \frac{1 - sin x}{1 + sin x}$ $RHS=(1+sinx)/(1-sinx)-(1-sinx)/(1+sinx)$

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

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

$=(2/(1-sinx)-cancel1-2/(1+sinx)+cancel1)$

$=2(1/(1-sinx)-1/(1+sinx))$

$=2((1+sinx-1+sinx)/((1-sinx)(1+sinx)))$

$=(4sinx)/(1-sin^2x)$

$=(4sinx)/cos^2x$

$=(4sinx)/cosx xx1/cosx$

$=4tanxsecx=LHS$

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How do you prove $4 tan x sec x = (\frac{1 + sin x}{1 - sin x}) - (\frac{1 - sin x}{1 + sin x})$ $4tanxsecx=((1+sinx)/(1-sinx))-((1-sinx)/(1+sinx))$ ?

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How do you prove $4 tan x sec x = (\frac{1 + sin x}{1 - sin x}) - (\frac{1 - sin x}{1 + sin x})$ $4tanxsecx=((1+sinx)/(1-sinx))-((1-sinx)/(1+sinx))$ ?