Question

${\tan }^{-1}\frac{1}{11}+{\tan }^{-1}\frac{2}{11}$ is equal to
(a) 0
(b) 1/2
(c) − 1
(d) none of these

Answer 1

(d) none of these

We know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$.
Now,
$$\begin{gathered} {\tan }^{-1}\frac{1}{11}+{\tan }^{-1}\frac{2}{11}={\tan }^{-1}\left(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{1}{11}\frac{2}{11}}\right) \\ ={\tan }^{-1}\left(\frac{\frac{3}{11}}{\frac{121-2}{121}}\right) \\ ={\tan }^{-1}\left(\frac{\frac{3}{11}}{\frac{119}{121}}\right) \\ ={\tan }^{-1}\left(\frac{33}{119}\right) \\ =0.27 \end{gathered}$$