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?

The formula
$$
\tan\frac{x}{2}=\pm\sqrt{\frac{1-x^2}{1+x^2}}
$$
is wrong. You were probably thinking to
$$
\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}
$$
but there's a better one that doesn't have any sign ambiguity:
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
If $x=\arcsin(5/13)$, then $\cos x>0$, so
$$
\cos x=\sqrt{1-\Bigl(\frac{5}{13}\Bigr)^2}=\frac{12}{13}
$$

As a side note: from
$$
\frac{1}{2}\arcsin\frac{5}{13}=\alpha
$$
you cannot argue that
$$
\sin\alpha=\frac{5}{26}
$$
You cannot bring $1/2$ inside the arcsine.
