How do you prove [(sinx-cosx)/(secx-cscx)]=[(sin2x)/2]?

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How do you prove [(sinx-cosx)/(secx-cscx)]=[(sin2x)/2]?

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Answer 1 · 2018-03-05T15:51:50

See below:

Explanation:

$sec x - csc x = \frac{1}{cos x} - \frac{1}{sin x} = \frac{sin x - cos x}{sin x cos x}$ $sec x - csc x = 1/cos x - 1/sin x = (sin x - cos x)/{sin x cos x}$
so that
$\frac{sin x - cos x}{sec x - csc x} = sin x cos x = \frac{1}{2} sin (2 x)$ $(sin x - cos x)/(sec x - csc x) = sin x cos x = 1/2sin(2x)$

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How do you prove [(sinx-cosx)/(secx-cscx)]=[(sin2x)/2]?

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