Properties of Inverse Trigonometric Functions – I
Property:
(i) \({{\tan }^{-1}}x+{{\tan }^{-1}}y=\left\{ \begin{align} & {{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\,if\,xy<1 \\ & \pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\,if\,x>0,y>0\,and\,xy>1 \\ & -\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right),\,if\,x<0,y>0\,and\,xy>1 \\ \end{align} \right.\).
(ii) \({{\tan }^{-1}}x-{{\tan }^{-1}}y=\left\{ \begin{align} & {{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right),\,if\,xy>-1 \\ & \pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right),\,if\,x>0,y>0\,and\,xy<-1 \\ & =\pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right),\,if\,x<0,y>0\,and\,xy<-1 \\ \end{align} \right.\).
Problem: Show that \({{\tan }^{-1}}\frac{1}{7}+{{\tan }^{-1}}\frac{1}{13}-{{\tan }^{-1}}\frac{2}{9}=0\).
Solution: Given that \({{\tan }^{-1}}\frac{1}{7}+{{\tan }^{-1}}\frac{1}{13}-{{\tan }^{-1}}\frac{2}{9}=0\).
Let us consider
\({{\tan }^{-1}}\frac{1}{7}=\alpha \),
\({{\tan }^{-1}}\frac{1}{13}=\beta \) and
\({{\tan }^{-1}}\frac{2}{9}=\gamma \),
\(\tan \alpha =\frac{1}{7}\),
\(\tan \beta =\frac{1}{13}\) and
\(\tan \gamma =\frac{2}{9}\),
\(\tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\),
\(\tan \left( \alpha +\beta \right)=\frac{\frac{1}{7}+\frac{1}{13}}{1-\left( \frac{1}{7}\times \frac{1}{13} \right)}\),
\(=\frac{\frac{20}{91}}{1-\frac{1}{91}}\),
= 20/ 90
tan (α + β) = 2/ 9
Now \(\tan \left( \alpha +\beta -\gamma \right)=\tan \left( \left( \alpha +\beta \right)-\gamma \right)=\frac{\tan \left( \alpha +\beta \right)-\tan \gamma }{1+\left( \tan \left( \alpha +\beta \right)\times \tan \gamma \right)}\),
\(\tan \left( \left( \alpha +\beta \right)-\gamma \right)=\frac{\frac{2}{9}-\frac{2}{9}}{1+\left( \frac{2}{9}\times \frac{2}{9} \right)}\),
tan ((α + β) – γ) = 0
(α + β) – γ = tan⁻¹ (0)
(α + β) – γ = tan⁻¹ (tan (0))
(∵ tan (tan⁻¹ (x)) = x)
(α + β) – γ = 0
Hence proved \({{\tan }^{-1}}\frac{1}{7}+{{\tan }^{-1}}\frac{1}{13}-{{\tan }^{-1}}\frac{2}{9}=0\).