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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 at 21:19

4 Answers 4

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

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Hint:

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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to complete the problem you need to know that $$\begin{array}{ll} 1)&\tan^2\theta+1=\sec^2\theta\\ 2)&\cos\theta=\frac{1}{\sec^\theta} \end{array}$$

The problem is actually one of motivation. The preceding identities are pretty basic and are one of the first ones learned (and reinforced through exercises). What is required is that the student does his homework, and reads his textbook. Too many students never read their textbook chapters and just pay attention (or not) to the teacher during class. How does a student motivate himself to behave like this? I have no idea.

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