Compute $2\arctan{2}+\arcsin{\frac{4}{5}}.$ Proceeding as you guys have tought me on my previous posts by setting the whole thing equal to $X$ taking the tangent of the whole expression and then using the additon formula for tangent: 
$$\tan{X}=\tan{\left(2\arctan{2}+\arcsin{\frac{4}{5}}\right)}=\frac{\tan{(2\arctan{2})}+\tan{\left(\arcsin{\frac{4}{5}}\right)}}{1-\tan{(2\arctan{2})}\cdot\tan{\left(\arcsin{\frac{4}{5}}\right)}}.$$
So lets split this into two sub-computations like this:
i): $$N=\tan{(2\arctan{2})}=\tan{(\arctan{2}+\arctan{2})}=\frac{2\tan{(\arctan{2})}}{1-\tan^2{(\arctan{2})}}=\frac{2\cdot 2}{1-2^2}=-\frac{4}{3}.$$
ii): 
$$M=\tan{\left(\arcsin{\frac{4}{5}}\right)}=\frac{\sin{\left(\arcsin{\frac{4}{5}}\right)}}{\cos{\left(\arcsin{\frac{4}{5}}\right)}}=\frac{\sin{\left(\arcsin{\frac{4}{5}}\right)}}{\sqrt{1-\sin^2{\left(\arcsin{\frac{4}{5}}\right)}}}=\frac{\frac{4}{5}}{\sqrt{1-\left(\frac{4}{5}\right)^2}}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}.$$
So,
$$\tan{X}=\frac{N+M}{1-MN}=\frac{-\frac{4}{3}+\frac{4}{3}}{1+\frac{4}{3}\cdot\frac{4}{3}}=0\Longrightarrow X=\arctan{0}=0.$$
The book says $X=\pi.$

The last implication is false: $\tan{X} = 0 \implies X = k\pi$ for some $k \in \mathbb{Z}$. To find out which value, it's probably simplest to estimate the terms: 


*

*$\arctan{2}>\arctan{1}=\pi/4$, so the sum is certainly bigger than $\pi/2$.

*On the other hand, $0<\arcsin{(4/5)} < \pi/2$ and $\arctan{2}<\pi/2$, so the result is bounded above by $3\pi/2$.


The only multiple of $\pi$ between these is $\pi$ itself, so the value of the sum must be $\pi$.
