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

i did the following working on LHS: $$\frac{\sin^2 x}{\tan x}=\frac{\sin(x)\sin(x)}{\tan x}=\sin(x)\cos(x)$$ i need to confirm that $$\frac{\sin x}{\tan x}=\cos(x)$$

Edward
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user118512
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    If you remember the definition of $\tan x$ in terms of $\sin x$ and $\cos x$, I think you'll be able to confirm your last step. – Edward Jan 14 '14 at 03:40

1 Answers1

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

user44197
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