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.