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

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Answer 1 · 2018-03-06T17:27:01

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$L H S$ $LHS$ : $\frac{cos x}{1 + sin x} + \frac{cos x}{1 - sin x}$ $cosx /(1+sinx) + cos x/(1-sinx)$

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

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

$=(cosxcancel(-sin x cos x)+cosx+cancel(sinx cos x))/cos^2x$

$=(2cosx)/cos^2x$

$=(2cancelcosx)/cos^cancel2x$

$=2/cosx$

$=2*1/cosx$

$=sec x$

$=RHS$

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