<s>[INST] 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?
 [/INST] Let $\arcsin(\frac{5}{13})=x$, thus $\sin(x)=\frac{5}{13}$
So, this problem reduced to find value of $\tan(\frac{x}{2})=u$,
According to $\sin(x)=\frac{2u}{u^2+1}$ identity for $u=\tan(\frac{x}{2})$;
$\frac{2u}{u^2+1}=\frac{5}{13}$
$5*(u^2+1)=13*2u$
$5u^2+5=26u$
$5u^2-26u+5=0$
$(u-5)*(5u-1)=0$
Due to $\tan(x)>0$ constraint, $\tan(\frac{x}{2})$ must be equal to $\frac{1}{5}$.
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