How do you prove that $\frac{{cos}^{2} x}{1 + sin 2 x} = \frac{1 - tan x}{1 + tan x}$ $(cos^2x)/(1 + sin2x) = (1 - tanx)/(1+ tanx)$ ?

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How do you prove that $\frac{{cos}^{2} x}{1 + sin 2 x} = \frac{1 - tan x}{1 + tan x}$ $(cos^2x)/(1 + sin2x) = (1 - tanx)/(1+ tanx)$ ?

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Answer 1 · 2016-05-01T01:37:12

Your identity is false!

Explanation:

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

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

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

Your identity is false, since the left side does not equal the right!!

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How do you prove that $\frac{{cos}^{2} x}{1 + sin 2 x} = \frac{1 - tan x}{1 + tan x}$ $(cos^2x)/(1 + sin2x) = (1 - tanx)/(1+ tanx)$ ?

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Explanation:

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How do you prove that $\frac{{cos}^{2} x}{1 + sin 2 x} = \frac{1 - tan x}{1 + tan x}$ $(cos^2x)/(1 + sin2x) = (1 - tanx)/(1+ tanx)$ ?