Question

Prove that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),\text{When}(xy<1)$.

Answer 1

Step 1:$\text{Let}{\tan }^{-1}x=A\text{and}{\tan }^{-1}y=B\text{. Then,}$

$\begin{array}{r}x=\tan A\text{and}y=\tan B\text{and}A,B\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\end{array}$

$\begin{array}{c}\therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\\ =\frac{x+y}{1-xy}\dots \left(1\right)\end{array}$

Now, the following cases arise.

Case 1) When $x>0,y>0$ and $xy<1$

In this case we have:

$x>0,y>0\mathrm{and}xy<1$

$\Rightarrow \frac{x+y}{1-xy}>0$

$\Rightarrow \tan (A+B)>0$ [Using (1)]

$\Rightarrow A+B$ lies in first or third quadrant.

$\Rightarrow 0<A+B<\frac{\mathrm{\pi }}{2}\left.\left[\begin{array}{l}\because x>0\Rightarrow 0<A<\frac{\mathrm{\pi }}{2}\\ y>0\Rightarrow 0<B<\frac{\mathrm{\pi }}{2}\end{array}\right\}\Rightarrow 0<A+B<\mathrm{\pi }\right]$

$\therefore \tan (A+B)=\frac{x+y}{1-xy}$ [From (1)]

$\Rightarrow A+B={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\left[\because 0<A+B<\frac{\mathrm{\pi }}{2}\right]$

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

Step 2: Case 2) When $x<0,y<0$ and $xy<1$

$\Rightarrow \frac{x+y}{1-xy}<0$

$\Rightarrow \tan (A+B)<0$ [From (1)]

$\Rightarrow A+B$ lies in second or fourth quadrant.

$\Rightarrow A+B$ lies in fourth quadrant

$\left.\left[\begin{array}{c}\because x<0\Rightarrow -\frac{\mathrm{\pi }}{2}<A<0\\ y<0\Rightarrow -\frac{\mathrm{\pi }}{2}<B<0\end{array}\right\}\Rightarrow -\mathrm{\pi }<A+B<0\right]$

$\Rightarrow -\frac{\mathrm{\pi }}{2}<A+B<0$

$\therefore \tan (A+B)=\frac{x+y}{1-xy}$

$\Rightarrow A+B={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

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

Hence, proved.