Suppose $\tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} \left( \frac{2x}{1 - x^2} \right)$,where $|x| < \frac{1}{\sqrt{3}}$,then one of the values of $y$ is
- A$\frac{3x + x^3}{1 + 3x^2}$
- B$\frac{3x - x^3}{1 + 3x^2}$
- C$\frac{3x + x^3}{1 - 3x^2}$
- D$\frac{3x - x^3}{1 - 3x^2}$
Explore More
Similar Questions
Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$,where $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ is equal to
DifficultJEE MAIN 2023
View SolutionThe solution set of the equation $\sin^{-1} x = 2\tan^{-1} x$ is
Medium
View SolutionIf $0 < x < 1$,then $\sqrt{1+x^2} [\{x \cos (\cot ^{-1} x)+\sin (\cot ^{-1} x)\}^2-1]^{\frac{1}{2}}$ is equal to
DifficultIIT 2008
View Solution$\cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$ is equal to
DifficultKCET 2013
View SolutionWith usual notations in $\Delta ABC$,if $C=90^{\circ}$,then $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo