સાબિત કરો કે: $\cot ^{2} \frac{\pi}{6} + \csc \frac{5 \pi}{6} + 3 \tan ^{2} \frac{\pi}{6} = 6$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $L.H.S. = \cot ^{2} \frac{\pi}{6} + \csc \frac{5 \pi}{6} + 3 \tan ^{2} \frac{\pi}{6}$
$= (\sqrt{3})^{2} + \csc \left(\pi - \frac{\pi}{6}\right) + 3 \left(\frac{1}{\sqrt{3}}\right)^{2}$
$= 3 + \csc \frac{\pi}{6} + 3 \times \frac{1}{3}$
$= 3 + 2 + 1 = 6$
$= R.H.S.$

Explore More

Similar Questions

જો $A = \sin 45^{\circ} + \cos 45^{\circ}$ અને $B = \sin 44^{\circ} + \cos 44^{\circ}$ હોય,તો

નીચેનામાંથી કયા ત્રિકોણમિતીય મૂલ્યો ઋણ છે?
$I. \sin(-292^{\circ})$
$II. \tan(-190^{\circ})$
$III. \cos(-207^{\circ})$
$IV. \cot(-222^{\circ})$

$\frac{\sqrt{3} \sin \theta + \cos \theta}{\sin \left(\theta + \frac{\pi}{6}\right)} = $

સાબિત કરો કે $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$.

જો $x \sin 45^\circ \cos^2 60^\circ = \frac{\tan^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$ હોય,તો $x = $

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