The value of $\tan \left( \tan^{-1} \frac{1}{2} - \tan^{-1} \frac{1}{3} \right)$ is

  • A
    $5/6$
  • B
    $7/6$
  • C
    $1/6$
  • D
    $1/7$

Explore More

Similar Questions

If $y = \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right)$,then $\frac{dy}{dx} = $

$\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) = . . . . .$

Considering the principal values of the inverse trigonometric functions,$\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right)$,for $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$,is equal to:

If $\alpha > \beta > \gamma > 0$,then the expression $\cot ^{-1}\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}$ is equal to:

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$,then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to

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