How to simplify $\frac{{sin}^{2} x}{{cos}^{2} + 3 cos x + 2} = \frac{1 - cos x}{2 + cos x}$ $sin^2x/(cos^2+3cosx+2)=(1-cosx)/(2+cosx)$ ?

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$\frac{{sin}^{2} X}{{cos}^{2} x + 3 cos X + 2} = \frac{1 - cos X}{2 + cos x}$ $(Sin^2X)/(cos^2x+3cosX+2)=(1-cosX)/(2+cosx)$

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Answer 1 · 2018-02-01T23:30:08

$L H S = \frac{{sin}^{2} x}{{cos}^{2} x + 3 cos x + 2}$ $LHS=sin^2x/(cos^2x+3cosx+2)$

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

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

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

$= \frac{1 - cos x}{2 + cos x} = R H S$ $=(1-cosx)/(2+cosx)=RHS$