How do you prove the identity $\frac{sin x - cos x}{sin x + cos x} = \frac{2 {sin}^{2} x - 1}{1 + 2 sin x cos x}$ $(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)$ ?

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How do you prove the identity $\frac{sin x - cos x}{sin x + cos x} = \frac{2 {sin}^{2} x - 1}{1 + 2 sin x cos x}$ $(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)$ ?

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Answer 1 · 2018-03-09T19:46:18

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Explanation:

Use the Property : ${sin}^{2} x + {cos}^{2} x = 1$ $color(blue)(sin^2x+cos^2x=1$

$L H S : \frac{sin x - cos x}{sin x + cos x}$ $LHS : (sinx-cosx)/(sinx+cosx)$

$= \frac{sin x - cos x}{sin x + cos x} \cdot \frac{sin x + cos x}{sin x + cos x}$ $=(sinx-cosx)/(sinx+cosx)* (sinx+cosx)/(sinx+cosx)$ -> multiply by conjugate

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

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

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

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

$= R H S$ $=RHS$

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How do you prove the identity $\frac{sin x - cos x}{sin x + cos x} = \frac{2 {sin}^{2} x - 1}{1 + 2 sin x cos x}$ $(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)$ ?

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How do you prove the identity $\frac{sin x - cos x}{sin x + cos x} = \frac{2 {sin}^{2} x - 1}{1 + 2 sin x cos x}$ $(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)$ ?