Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of sum and difference of two and three angles
Mediumसिद्ध कीजिए कि $\sin^{2} 6x - \sin^{2} 4x = \sin 2x \sin 10x$.
(N/A) हम सर्वसमिका $\sin^{2} A - \sin^{2} B = \sin(A+B) \sin(A-B)$ का उपयोग करते हैं।
$L.H.S. = \sin^{2} 6x - \sin^{2} 4x$
$= \sin(6x + 4x) \sin(6x - 4x)$
$= \sin(10x) \sin(2x)$
$= \sin 2x \sin 10x$
$= R.H.S.$
$L.H.S. = \sin^{2} 6x - \sin^{2} 4x$
$= \sin(6x + 4x) \sin(6x - 4x)$
$= \sin(10x) \sin(2x)$
$= \sin 2x \sin 10x$
$= R.H.S.$
Explore More
Similar Questions
यदि $b \sin \alpha = a \sin (\alpha + 2\beta)$ है,तो $\frac{a + b}{a - b} = $
Easy
View Solution$\cos \frac{\pi}{12} = ?$
$\frac{\tan 70^o - \tan 20^o}{\tan 50^o}$ का मान ज्ञात कीजिए।
Medium
View Solutionसिद्ध कीजिए कि $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$
Medium
View Solutionयदि $\tan \theta = \frac{\cos 25^{\circ} + \sin 25^{\circ}}{\cos 25^{\circ} - \sin 25^{\circ}}$ और $\theta$ तीसरे चतुर्थांश में है,तो $\theta =$ ($^{\circ}$ में)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo