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