\(tan^{-1}1+tan^{-1}\frac{1}{3}=?\)A. \(tan^{-1}\frac{4}{3}\)B. \(ta

1.	\(tan^{-1}1+tan^{-1}\frac{1}{3}=?\)A. \(tan^{-1}\frac{4}{3}\)B. \(tan^{-1}\frac{2}{3}\)C. \(tan^{-1}2\)D. \(tan^{-1}3\)
Answer» Correct Answer is (C) tan^-12 Let , x = \(tan^{-1}1+tan^{-1}\frac{1}{3}\) Since we know that tan^-1x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\) ⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)= tan^-1\((\frac{1+\frac{1}{3}}{1-\frac{1}{3}})\)= \(tan^{-1}2\)

\(tan^{-1}1+tan^{-1}\frac{1}{3}=?\)A. \(tan^{-1}\frac{4}{3}\)B. \(tan^{-1}\frac{2}{3}\)C. \(tan^{-1}2\)D. \(tan^{-1}3\)

Answer»

Correct Answer is (C) tan^-12

Let , x = \(tan^{-1}1+tan^{-1}\frac{1}{3}\)

Since we know that tan^-1x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\)

⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)= tan^-1\((\frac{1+\frac{1}{3}}{1-\frac{1}{3}})\)= \(tan^{-1}2\)