How do you prove $\frac{cos x}{1 + sin x} - 2 sec x = \frac{cos x}{sin x - 1}$ $cosx/(1+sinx)-2secx=cosx/(sinx-1)$ ?

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Answer 1 · 2016-04-17T00:20:05

See below

LHS =left hand side, RHS =right hand side

LHS $= (\frac{cos x}{1 + sin x}) \times \frac{1 - sin x}{1 - sin x} - \frac{2}{cos x}$ $=(cosx/(1+sinx)) xx (1-sinx)/(1-sinx) - 2/cosx$

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

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

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

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

$= \frac{- 1 - sin x}{cos x}$ $=(-1-sinx)/cosx$

$= - \frac{1 + sin x}{cos x} \times \frac{1 - sin x}{1 - sin x}$ $=-(1+sinx)/cosx xx (1-sinx)/(1-sinx)$

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

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

$= \frac{cos x}{- (1 - sin x)}$ $= cosx/ (-(1-sinx))$

$= \frac{cos x}{sin x - 1}$ $=cosx/(sinx-1)$

$= R H S$ $=RHS$

How do you prove cosx/(1+sinx)-2secx=cosx/(sinx-1)cosx1+sinx−2secx=cosxsinx−1cosx/(1+sinx)-2secx=cosx/(sinx-1)?