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I'am trying to prove the below identity, but this is what i end up getting. $$\begin{align*} \frac{2\tan(x)}{1 + \tan^2(x)} &= \sin(2x)\\ &= \frac{2\frac{\sin(x)}{\cos(x)}}{1 + \frac{\sin^2(x)}{\cos^2(x)}}\\ &= 2\sin(x) \end{align*}$$

Can someone point out the mistake for me please?

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$\frac{\cos^2 x}{\cos x} = \cos x \neq 1$ for general $x$. –  Pacciu Mar 14 '12 at 18:36
    
Your last step went wrong. It should be $\frac{2\frac{\sin(x)}{\cos(x)}}{1 + \frac{\sin^2(x)}{\cos^2(x)}} = 2\frac{\frac{\sin(x)}{\cos(x)}}{{\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}}} = 2\sin(x)\cos(x) = \sin(2x)$ Using the fact that $\sin^2(x)+\cos^2(x)=1$ –  Kirthi Raman Mar 14 '12 at 20:30

3 Answers 3

up vote 2 down vote accepted

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

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Hint: $1=\frac{cos^2(x)}{cos^2(x)}$

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after solving the middle step, you will end up as 1/cos^2(x) as denominator. After dividing, it becomes 2*sin(x)*cos(x) which is equal to sin(2x).

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