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I am sorry to ask so many of these questions in such as short time span.

But how would I prove this following trigonometric identity. $ \frac{1+\cos(2A)}{\sin(2A)}=\cot A $ My work thus far is $ \frac{1+\cos^2A-\sin^2A}{2\sin A\cos A} $ I know $1-\sin^2A=\cos^2A$

So I do $ \frac{\cos^2A+\cos^2A}{2\sin A\cos A} $ I know not what I do next.

  • 0
    There's also a nice way of seeing this. Take a look at [this answer](http://math.stackexchange.com/questions/120704/how-to-prove-a-trigonometric-identity/126075#126075) by robjohn.2012-07-26

3 Answers 3

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$ \begin{align} \frac{1+\cos(2A)}{\sin(2A)} &=\frac{1+\cos^2(A)-\sin^2(A)}{2\sin(A)\cos(A)}\tag{1}\\ &=\frac{\csc^2(A)+\cot^2(A)-1}{2\,\cot(A)}\tag{2}\\ &=\frac{2\,\cot^2(A)}{2\,\cot(A)}\tag{3}\\[4pt] &=\cot(A)\tag{4} \end{align} $

  1. double angle formulas

  2. multiply numerator and denominator by $\csc^2(A)$

  3. $\cot^2(A)+1=\csc^2(A)$

  4. cancel $2\cot(A)$ in numerator and denominator

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$\frac{1+\cos(2A)}{\sin(2A)} = \frac{2\cos^2 A}{2\sin A \cdot \cos A} = \frac{\cos A}{\sin A} = \cot A$

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    thanks @mixedmath. should I write the answer if someone already give2013-05-09
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Just write $\cos^2 A+\cos^2 A=2\cos^2 A$ (a quantity added to itself is twice the quantity). Then write $2\cos^2 A=2\cos A\cdot\cos A$ and cancel a $2\cos A$ term in the numerator with the $2\cos A$ term in the denominator.