How do we prove the trigonometric identity $(1 - \cos x)\left(1 + \frac{1}{\cos x}\right) = \tan x\sin x$?

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Please show me the steps to completing this: $$(1-\cos x)\left(1+\frac{1}{\cos x}\right) = \tan x\sin x$$

Thanks.

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3 Answers

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$(1-\cos x)(1+\frac{1}{\cos x})$

$=1+\frac{1}{\cos x}-\cos x-1$

$=\frac{1}{\cos x}-\cos x$

$=\frac{1-\cos^2 x}{\cos x}$

$=\frac{\sin^2 x}{\cos x}$

$=\sin x\tan x$

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$$(1-\cos x)(\frac{1+\cos x}{\cos x})=\frac{1-\cos^2x}{\cos x}=\frac{\sin x\cdot \sin x}{\cos x}= \tan x \cdot\sin x$$

That's all.

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Notice, $$(1-\cos x)\left(1+\frac{1}{\cos x}\right)$$ $$=(1-\cos x)\left(1+\sec x\right)$$ $$=1+\sec x-\cos x-1$$ $$=\frac{1}{\cos x}-\cos x$$ $$=\frac{1-\cos^2 x}{\cos x}$$ $$=\frac{\sin^2 x}{\cos x}=\sin x\tan x$$

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