$\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ} = $

  • A
    $1 - 2 \sin 10^{\circ} \sin 15^{\circ} \sin 20^{\circ}$
  • B
    $4 \cos 20^{\circ} \cos 30^{\circ} \cos 40^{\circ}$
  • C
    $4 \cos 10^{\circ} \cos 15^{\circ} \cos 20^{\circ}$
  • D
    $4 \cos 25^{\circ} \cos 30^{\circ} \cos 35^{\circ}$

Explore More

Similar Questions

यदि $\tan A=\frac{1}{\sqrt{x(x^2+x+1)}}, \tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ और $\tan C=\sqrt{x^{-1}+x^{-2}+x^{-3}}$ है,तो:

यदि $\cos (\alpha + \beta) = \frac{3}{5}$,$\sin (\alpha - \beta) = \frac{5}{13}$ और $0 < \alpha, \beta < \frac{\pi}{4}$ है,तो $\tan (2\alpha)$ का मान ज्ञात कीजिए।

यदि $\frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}$ है,तो $\frac{\tan x}{\tan y} = $

$\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ का मान है

यदि $\tan A = -\frac{1}{2}$ और $\tan B = -\frac{1}{3}$ है,तो $A + B = $

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