Question #6a568

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Answer 1 · 2018-02-19T16:12:24

I guess you want me to simplify

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

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

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

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

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

$\frac{{sin}^{2} x (1 - csc x)}{- {cos}^{2} x} =$ $(sin^2x(1-cscx))/(-cos^2x)=$

$- {tan}^{2} x (1 - csc x) =$ $-tan^2x(1-cscx)=$

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

$- {tan}^{2} x + tan x \cdot sec x =$ $-tan^2x+tanx*secx=$

$- tan x (tan x - sec x)$ $-tanx(tanx-secx)$