# Sum of Angles in Terms of tan⁻¹ – Part2

#### Sum of Angles in Terms of tan⁻¹ – Part2

{{\tan }^{-1}}x+{{\tan }^{-1}}y=\left\{ \begin{align} & \pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\ \ \ if\ \ x>0,y>0,\ and\ xy>1 \\ & -\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\ \ \ if\ \ x<0,y<0,\ and\ xy>1 \\ \end{align} \right.

Proof:

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

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

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

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

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

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

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

From the graph, we have

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

=\left\{ \begin{align} & \alpha ,\ \ \ \ \ \ \ \left( -\pi /2 \right)\le \alpha \le \left( \pi /2 \right) \\ & \alpha -\pi ,\ \ \left( -\pi /2 \right)<\alpha <\pi \\\end{align} \right.,

=\left\{ \begin{align}& {{\tan }^{-1}}x+{{\tan }^{-1}}y,\ \ \ \ \ \ \ \left( -\pi /2 \right)\le {{\tan }^{- 1}}x+{{\tan }^{-1}}y\le \left( \pi /2 \right) \\& {{\tan }^{-1}}x+{{\tan }^{-1}}y-\pi ,\ \ \left( -\pi /2 \right)<{{\tan }^{-1}}x+{{\tan }^{1}}y<\pi\\\end{align} \right.,

Case(i):

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

⇒ xy > 1

$${{\tan }^{-1}}x>\pi /2-{{\tan }^{-1}}y$$,

$${{\tan }^{-1}}x>{{\cot }^{-1}}y$$,

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

⇒ x > (1/y)

⇒xy>1

Case(ii):

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

⇒ xy > 1