The number of solutions of $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$ is

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $> 2$

Explore More

Similar Questions

If $y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$,then $\frac{dy}{dx} = $

If $y = \tan^{-1}\left(\frac{1}{1+x+x^2}\right) + \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) + \tan^{-1}\left(\frac{1}{x^2+5x+7}\right)$,then the value of $y'(0)$ is

If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive integers,then the possible values of $(x, y)$ are:

Difficult
View Solution

Match List $I$ with List $II$ and select the correct answer using the code given below the lists:
List $I$ List $II$
$P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value $1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$
$Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is $2$. $\sqrt{2}$
$R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is $3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$, then possible value of $x$ is $4$. $1$

Codes: $P \quad Q \quad R \quad S$

The value of $\lim_{x \to 1/\sqrt{2}} \frac{x - \cos(\sin^{-1} x)}{1 - \tan(\sin^{-1} x)}$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo