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

1.	\(tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}\)=?A. \(\frac{\pi}{3}\)B. \(\frac{\pi}{4}\) C. \(\frac{\pi}{2}\) D. \(\frac{2\pi}{3}\)
Answer» Correct Answer is (B) \(\frac{\pi}{4}\) Let , x = \(tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}\) 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\((\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{3}\times\frac{2}{3})})\)= \(tan^{-1}1\) = \(\frac{\pi}{4}\)

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

Answer»

Correct Answer is (B) \(\frac{\pi}{4}\)

Let , x = \(tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}\)

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\((\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{3}\times\frac{2}{3})})\)= \(tan^{-1}1\) = \(\frac{\pi}{4}\)