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Verify this equation by changing only one side:

$$\frac{2\cot(\theta)}{\cot(\theta)+\tan(\theta)}=2\cos^2(\theta)$$ Please show your work.`

I multiplied the numerator and denominator by $\tan(\theta)$ and got $\dfrac{2}{1+tan^2(\theta)}$ but now I'm stuck. Could someone help me out?

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Yeah you're right it is sorry about that. – Nick Jan 27 '14 at 21:19

$$ \frac{2 \cot(t) }{ \cot(t) + \tan(t) } = \frac{2 \frac{\cos(t)}{\sin(t)} }{ \frac{\cos(t)}{\sin(t)} + \frac{\sin(t)}{\cos(t)} } = \frac{2 \cos^2(t)}{ \cos^2(t) + \sin^2(t) } = 2 \cos^2(t) . $$

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Next step: Multiply by $\dfrac{\cos^2(\theta)}{\cos^2(\theta)}$ and use the identity $\sin^2(\theta)+\cos^2(\theta)=1$

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To continue your solution, recognize $1+\tan^2\theta = \sec^2\theta$; then the remaining task is to get from $\frac2{\sec^2\theta}$ to $2\cos^2\theta$.

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