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

1.	2 tan-1\(\frac{1}{3}\)= ?A. tan-1\(\frac{3}{2}\)B. tan-1\(\frac{3}{4}\)C. tan-1\(\frac{4}{3}\)D. none of these
Answer» Correct Answer is (B) tan^-1\(\frac{3}{4}\) Let , x = tan^-1\(\frac{1}{3}\) + 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\((\frac{\frac{1}{3}+\frac{1}{3}}{1-(\frac{1}{3}\times\frac{1}{3})})\)= \(tan^{-1}1\)\(\frac{3}{4}\)

2 tan-1\(\frac{1}{3}\)= ?A. tan-1\(\frac{3}{2}\)B. tan-1\(\frac{3}{4}\)C. tan-1\(\frac{4}{3}\)D. none of these

Answer»

Correct Answer is (B) tan^-1\(\frac{3}{4}\)

Let , x = tan^-1\(\frac{1}{3}\) + 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\((\frac{\frac{1}{3}+\frac{1}{3}}{1-(\frac{1}{3}\times\frac{1}{3})})\)= \(tan^{-1}1\)\(\frac{3}{4}\)