Find $\tan(\frac{1}{2}\arcsin(\frac{5}{13}))$ I wanted to solve it by using this formula: $\tan(\frac{x}{2})=\pm\sqrt{\frac{1-x^2}{1+x^2}}.$ I thought it wouldn't work (because there are $\pm$). Then used the right triangle method: $$\frac{1}{2}\arcsin(\frac{5}{13})=\alpha\Rightarrow\frac{5}{26}=\sin\alpha$$ $$a^2+b^2=c^2\Rightarrow a^2+25=676\Rightarrow a=\sqrt{651}\Rightarrow\tan(\frac{5}{\sqrt{651}}).$$ It turned out to be wrong. How to get the right answer?

I think you need to check you first formula. You can use $\arcsin (\frac {2x}{1+x^2})=2\arctan (x) $ and $\arctan (\frac {2x}{1-x^2})=2\arctan (x) $. Both of which can be proved using $x=\tan (t)$. I think you can continue from here.
