How do you simplify $\frac{sin x}{1 + sin x} - \frac{sin x}{1 - sin x}$ $sinx/(1 + sinx) - sinx/(1 -sinx)$ ?

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Answer 1 · 2016-04-17T01:46:29

You should get $- 2 {tan}^{2} x$ $-2tan^2x$ as a final answer.

$\frac{sin x}{1 + sin x} - \frac{sin x}{1 - sin x}$ $(sinx)/(1 + sinx) - sinx/(1 - sinx)$

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

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

Using the pythagorean identity ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$

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

Using the quotient identity $tan x = \frac{sin x}{cos x}$ $tanx = sinx/cosx$

$= - 2 {tan}^{2} x$ $=-2tan^2x$

Hopefully this helps!

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