Question #9a883

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Question #9a883

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Answer 1 · 2017-02-10T03:31:36

This proof actually doesn't require more than a couple steps.

Explanation:

$sec x + csc x = \frac{1}{cos x} + \frac{1}{sin x}$ $secx+cscx=1/cosx+1/sinx$

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

$sec x csc x = \frac{1}{sin x cos x}$ $secxcscx=1/(sinxcosx)$

$\frac{sec x + csc x}{sec x csc x} = \frac{\frac{sin x + cos x}{sin x cos x}}{\frac{1}{sin x cos x}}$ $(secx+cscx)/(secxcscx)=((sinx+cosx)/(sinxcosx))/(1/(sinxcosx))$

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

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Question #9a883

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