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How do you prove (tanx-cotx)/(sinxcosx)=sec^2(x)-csc^2(x)?

1 Answer

Jan 9, 2018

See below

Explanation:

$\frac{tan x - cot x}{sin x cos x}$ $(tanx-cotx)/(sinxcosx)$

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

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

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

$= {sec}^{2} x - {csc}^{2} x$ $=sec^2x-csc^2x$ , as required. $square$

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