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Verify

$$\frac{\cot x -\tan x}{\cos x + \sin x}=\frac{\cos x - \sin x}{\sin x \cos x}$$

After several tries I cannot find a concrete way of proving/verifying this. Any help/hints?

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$$\frac{\cot x-\tan x}{\cos x+\sin x}=\frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{\cos x+\sin x}=\frac{\cos^2x-\sin^2x}{\cos x\sin x(\cos x+\sin x)}=$$

$$=\frac{\cos x-\sin x}{\cos x\sin x}$$

Using from first line to second the difference of squares identity:

$$a^2-b^2=(a-b)(a+b)$$

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  • 1
    $\begingroup$ Very quick response and brilliantly done! $\endgroup$ – EconDude May 16 '16 at 22:33

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