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

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Answer 1 · 2018-03-17T05:27:33

Please refer to the Explanation.

We have, $\frac{cos x}{1 - sin x}$ $cosx/(1-sinx)$ ,

$= \frac{cos x}{1 - sin x} \times 1$ $=cosx/(1-sinx)xx1$ ,

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

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

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

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

$= \frac{1}{cos x} + \frac{sin x}{cos x}$ $=1/cosx+sinx/cosx$ ,

$= sec x + tan x$ $=secx+tanx$ , as desired!

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How do you prove cos(x)1−sin(x)=sec(x)+tan(x)cos(x) / (1-sin(x)) = sec(x) + tan(x)?