Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$ 
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.

My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}})+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{63}{-16}+\tan^{-1}\frac{63}{16}$$
$$=-\tan^{-1}\frac{63}{16}+\tan^{-1}\frac{63}{16}$$
$$=0$$
But the answer is given as $\pi$. What is my mistake?

Now I am going to prove that:
$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$
for all $x,y\in\left]0,+\infty\right[$ such that $xy>1$.
Proof:
For all  $x,y\in\left]0,+\infty\right[$ such that $xy>1$, it results that:
$x>\frac{1}{y}=\cot\left(\tan^{-1}y\right)=\tan\left(\frac{\pi}{2}-\tan^{-1}y\right)$.
Since $\frac{\pi}{2}-\tan^{-1}y\in \left]0,\frac{\pi}{2}\right[$ and tangent is an invertible and monotone increasing function in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that:
$\tan^{-1}x>\frac{\pi}{2}-\tan^{-1}y$, therefore:
$\tan^{-1}x+\tan^{-1}y-\pi>-\frac{\pi}{2}$. (*)
On the other hand,
$\tan^{-1}x+\tan^{-1}y<\pi$, therefore:
$\tan^{-1}x+\tan^{-1}y-\pi<0$. (**)
From (*) and (**), it follows that:
$\tan^{-1}x+\tan^{-1}y-\pi \in \left]-\frac{\pi}{2},0\right[$.
$\tan\left(\tan^{-1}x+\tan^{-1}y-\pi\right)=\frac{\tan\left(\tan^{-1}x\right)+ \tan\left(\tan^{-1}y-\pi\right)}{1- \tan\left(\tan^{-1}x\right)\cdot\tan\left(\tan^{-1}y-\pi\right)}$
$\tan\left(\tan^{-1}x+\tan^{-1}y-\pi\right)=\frac{\tan\left(\tan^{-1}x\right)+ \tan\left(\tan^{-1}y\right)}{1- \tan\left(\tan^{-1}x\right)\cdot\tan\left(\tan^{-1}y\right)}$
$\tan\left(\tan^{-1}x+\tan^{-1}y-\pi\right)=\frac{x+y}{1-xy}$
Since $\tan^{-1}x+\tan^{-1}y-\pi \in \left]-\frac{\pi}{2},0\right[$ and tangent is invertible in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that:
$\tan^{-1}x+\tan^{-1}y-\pi=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, therefore:
$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$.
