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

Question

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?

Answer 1

Actually

$\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}=$

$=\pi+\tan^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}\right)$

We can notice that

$\frac{\pi}{2}<\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}<\pi$.

I am going to prove that

if $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{2}$ then $\alpha+\beta=\pi+\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right).$

Proof:

$\tan(\alpha+\beta-\pi)=\frac{\tan\alpha+\tan(\beta-\pi)}{1-\tan\alpha\cdot\tan(\beta-\pi)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}$

As $|\alpha+\beta-\pi|<\frac{\pi}{2}$ and the function tangent is invertible in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that

$\alpha+\beta-\pi=\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right)$, therefore:

$\alpha+\beta=\pi+\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right)$.

Answer 2

After seeing Angelo's answer, I realized my mistake. So, posting the following for my future reference and for others to see.

$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac{x+y}{1-xy}), xy\lt 1$$

$$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}(\frac{x+y}{1-xy}), xy\gt 1;x,y\gt0$$

Answer 3

Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

As $\dfrac{12}5\cdot\dfrac{3}4>1,$

$$\tan^{-1}\dfrac{12}5+\tan^{-1}\dfrac{3}4=\pi+\tan^{-1}\left(\dfrac{\dfrac{12}5+\dfrac{3}4}{1-\dfrac{12}5\cdot\dfrac{3}4}\right)=?$$

$$\text{and }\tan^{-1}(-y)=-\tan^{-1}y$$

Alternatively,

$$(5+12i)(4+3i)(16+63i)=(36i-16)(36i+16)=-(36)^2-16^2$$

Apply argument in both sides.

Now as $0<\tan^{-1}\dfrac{12}5,\tan^{-1}\dfrac34,\tan^{-1}\dfrac{63}{16}<\dfrac\pi2,$

$$0<\tan^{-1}\dfrac{12}5+\tan^{-1}\dfrac34+\tan^{-1}\dfrac{63}{16}<\dfrac{3\pi}2$$

Answer 4

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:

$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$.