Question #01cbb

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Answer 1 · 2017-03-01T09:41:43

$\frac{cos x}{1 + sin x} + tan x$ $cosx/(1+sinx)+tanx$

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

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

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

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

$= \frac{1}{cos x} - \frac{sin x}{cos x} + tan x$ $=1/cosx-sinx/cosx+tanx$

$= \frac{1}{cos x} - tan x + tan x$ $=1/cosx-tanx+tanx$

$= \frac{1}{cos x}$ $=1/cosx$