Find the value of $3\tan^{-1}\left(\dfrac{1}{2}\right)+2\tan^{-1}\left(\dfrac{1}{5}\right)+\sin^{-1}\left(\dfrac{142}{65\sqrt{5}}\right)$
My reference gives the solution $0$ to this problem.
My Attempt
$$ |x_1|=\frac{1}{2}\leq\frac{1}{\sqrt{3}}\implies 3\tan^{-1}\frac{1}{2}=\tan^{-1}\frac{\frac{3}{2}-\frac{1}{8}}{1-\frac{3}{4}}=\tan^{-1}\frac{11}{2}\\ |x_2|=\frac{1}{5}<1\implies2\tan^{-1}\dfrac{1}{5}=\tan^{-1}\frac{\frac{2}{5}}{1-\frac{1}{25}}=\tan^{-1}\frac{10}{24}=\tan^{-1}\frac{5}{12}\\ XY=\frac{11}{2}\frac{5}{12}=\frac{55}{24}>1\quad\&\quad X,Y>0\\ 3\tan^{-1}\dfrac{1}{2}+2\tan^{-1}\dfrac{1}{5}=\tan^{-1}\frac{11}{2}+\tan^{-1}\frac{5}{12}=\pi+\tan^{-1}\frac{\frac{11}{2}+\frac{5}{12}}{1-\frac{11}{2}\frac{5}{12}}\\ =\pi-\tan^{-1}\frac{142}{31} $$ So I seem to get $\pi$ as the answer, what is going wrong here ?
Your reference is clearly wrong without calculating anything. Functions like $\tan^{-1}$ and $\sin^{-1}$ generate positive values for positive arguments, by convention. All your arguments are positive so the result must be positive too. Numeric check proves that your result is correct.