Question #849cc

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Question #849cc

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Answer 1 · 2016-10-18T22:19:12

$\frac{1}{sin x cos x} - \frac{sin x}{cos x} = cot x$ $1/(sinxcosx)-sinx/cosx=cotx$

Explanation:

$\frac{1}{sin x cos x} - \frac{sin x}{cos x}$ $1/(sinxcosx)-sinx/cosx$ = $\frac{1}{cos x} (\frac{1}{sin x} - sin x)$ $1/cosx(1/sinx-sinx)$

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

But $1 - {sin}^{2} x = {cos}^{2} x$ $1-sin^2x=cos^2x$

So the expression can be written as
$\frac{{cos}^{2} x}{cos x sin x}$ $cos^2x/(cosxsinx)$

Cancelling gives
$\frac{cos x}{sin x}$ $cosx/sinx$ = $cot x$ $cotx$

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Question #849cc

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