prove that:
1. $\displaystyle \frac{\sin A + \sin 2A}{\cos A - \cos 2A} =\cot (A/2)$
2.$\displaystyle \frac{1+\cos A}{\sin A} = \cot (A/2) $
please help...
For the first one I think it can be useful to work with the identities:
$\displaystyle \sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)$$\displaystyle \cos\left(\frac{x-y}{2}\right)$
$\displaystyle \cos(x)-\cos(y)=-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$
Hello, earthboy!
$\displaystyle \text{Prove that: }\:\frac{1+\cos A}{\sin A} \:=\: \cot \frac{A}{2}$
Right side: .$\displaystyle \cot\frac{A}{2} \;=\;\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} \;=\;\frac{\sqrt{\frac{1+\cos A}{2}}}{\sqrt{\frac{1-\cos A}{2}}} \;=\;\sqrt{\frac{1+\cos A}{1-\cos A}}$
Multiply by $\displaystyle \tfrac{1+\cos A}{1+\cos A}\!:\;\;\sqrt{\frac{1+\cos A}{1-\cos A}\cdot\frac{1+\cos A}{1+\cos A}} \;=\;\sqrt{\frac{(1+\cos A)^2}{1-\cos^2A}}$
. . . . . . . . . . . . $\displaystyle =\;\sqrt{\frac{(1+\cos A)^2}{\sin^2\!A}} \;=\; \frac{1+\cos A}{\sin A}$