સાબિત કરો કે: $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) આપણે ત્રિકોણમિતીય નિત્યસમનો ઉપયોગ કરીએ છીએ:
$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$L.H.S.$ પર આ લાગુ પાડતા:
$L.H.S. = \frac{2 \sin \left( \frac{5x+3x}{2} \right) \cos \left( \frac{5x-3x}{2} \right)}{2 \cos \left( \frac{5x+3x}{2} \right) \cos \left( \frac{5x-3x}{2} \right)}$
પદોનું સાદું રૂપ આપતા:
$= \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}$
સામાન્ય પદો $2$ અને $\cos(x)$ ને દૂર કરતા:
$= \frac{\sin(4x)}{\cos(4x)} = \tan(4x) = R.H.S.$
આમ,નિત્યસમ સાબિત થાય છે.

Explore More

Similar Questions

$\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \cos(54^{\circ}+A) \cos(54^{\circ}-A) = $

સાબિત કરો કે: $\sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

Difficult
View Solution

$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$ ની કિંમત શોધો.

સાબિત કરો કે: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$

Difficult
View Solution

જો $(1 + \tan \theta )(1 + \tan \phi ) = 2$ હોય,તો $\theta + \phi = \dots ^\circ$

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