If \(tan^{-1}x=\frac{\pi}{4}-tan^{-1}\frac{1}{3}\) then x = ?A. \(\f

1.	If \(tan^{-1}x=\frac{\pi}{4}-tan^{-1}\frac{1}{3}\) then x = ?A. \(\frac{1}{2}\) B. \(\frac{1}{4}\)C. \(\frac{1}{6}\)D. none of these
Answer» Correct Answer is (A) \(\frac{1}{2}\) Now, tan^-1 x = tan^-1 1 - tan-1 \(\frac{1}{3}\) \((\because tan\frac{\pi}{4}=1)\) 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\((\frac{1-\frac{1}{3}}{1+\frac{1}{3}})\)= \(tan^{-1}1\)\(\frac{1}{2}\) ⇒ tan^-1 x = \(tan^{-1}1\)\(\frac{1}{2}\) ⇒ x = \(\frac{1}{2}\)

If \(tan^{-1}x=\frac{\pi}{4}-tan^{-1}\frac{1}{3}\) then x = ?A. \(\frac{1}{2}\) B. \(\frac{1}{4}\)C. \(\frac{1}{6}\)D. none of these

Answer»

Correct Answer is (A) \(\frac{1}{2}\)

Now, tan^-1 x = tan^-1 1 - tan-1 \(\frac{1}{3}\) \((\because tan\frac{\pi}{4}=1)\)

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

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

⇒ x = \(\frac{1}{2}\)