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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.

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$\cos^2 A+\cos^2 A=2\cos^2 A$. – David Mitra Jul 26 '12 at 17:54
$\frac{cos^2A+cos^2A}{2\sin A\cos A}=\frac{2\cos^2A}{2\sin A \cos A}=\cot A$ – Saurabh Jul 26 '12 at 17:56
There's also a nice way of seeing this. Take a look at this answer by robjohn. – Ian Mateus Jul 26 '12 at 18:24

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

  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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I have rolled back to TMM's edit, since it is exceptionally clear and easy to read. – mixedmath May 9 '13 at 19:17
why is this downvoted – mathguy May 9 '13 at 19:19
thanks @mixedmath. should I write the answer if someone already give – iostream007 May 9 '13 at 19:42

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.

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Thank you your math skills are excellent. – Fernando Martinez Jul 26 '12 at 18:00

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