# Sum of Angles in Terms of tan⁻¹ – Part1

## Sum of Angles in Terms of tan⁻¹ – Part1

Theorem:  $${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\ \ \ if\ \ xy<1$$,

Proof:

Let $${{\tan }^{-1}}x=A$$ and $${{\tan }^{-1}}y=B$$,

Where $$A,B\in \left( -\pi /2,\pi /2 \right)$$,

$$\tan \left( A+B \right)=\frac{\operatorname{Tan}A+\tan B}{1-\tan A\tan B}$$,

$$=\frac{x+y}{1-xy}$$,

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

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

From graph

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

$$=\alpha +\pi ,\ \ -\pi <\alpha <\left( -\pi /2 \right)$$,

Where $$\alpha \in \left( -\pi,\pi \right)$$,

= $${{\tan }^{-1}}x+{{\tan }^{-1}}y+\pi ,\ \ \ \ \ -\pi <{{\tan }^{-1}}x+{{\tan }^{-1}}y<\left( -\pi /2 \right)$$,

$$-\pi <{{\tan }^{-1}}x+{{\tan }^{-1}}y<\left( -\pi /2 \right)$$,

⇒ x<0, y<0,

$${{\tan }^{-1}}x<\left( -\pi /2 \right)-{{\tan }^{-1}}y$$,

$${{\tan }^{-1}}x<-\left( \pi /2 \right)-{{\tan }^{-1}}\left( -y \right)$$,

$$\Rightarrow {{\tan }^{-1}}x<{{\tan }^{-1}}\left( \frac{1}{y} \right)$$,

⇒ x<(1/y),

⇒xy<1.