$\tan\, 20^{\circ} + \tan\, 40^{\circ} + \sqrt{3}\, \tan\, 20^{\circ} \tan\, 40^{\circ}$ is equal to

  • A
    $\frac{\sqrt{3}}{2}$
  • B
    $\frac{\sqrt{3}}{4}$
  • C
    $\sqrt{3}$
  • D
    $1$

Explore More

Similar Questions

$\cos 48^{\circ} \cdot \cos 12^{\circ} = ?$

If $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$,where $A$ and $B$ lie in the first and third quadrant respectively,then $\cos(A + B) = $

Prove that $\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)$

The value of $\frac{\sin \theta + \sin 3 \theta}{\cos \theta + \cos 3 \theta}$ is

If $\cot \alpha = 1$ and $\sec \beta = -\frac{5}{3}$,where $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$,then the value of $\tan(\alpha + \beta)$ and the quadrant in which $\alpha + \beta$ lies,respectively,are

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