Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of sum and difference of two and three angles
Mediumસાબિત કરો કે $\sin^{2} 6x - \sin^{2} 4x = \sin 2x \sin 10x$.
(N/A) આપણે નિત્યસમ $\sin^{2} A - \sin^{2} B = \sin(A+B) \sin(A-B)$ નો ઉપયોગ કરીએ છીએ.
$L.H.S. = \sin^{2} 6x - \sin^{2} 4x$
$= \sin(6x + 4x) \sin(6x - 4x)$
$= \sin(10x) \sin(2x)$
$= \sin 2x \sin 10x$
$= R.H.S.$
$L.H.S. = \sin^{2} 6x - \sin^{2} 4x$
$= \sin(6x + 4x) \sin(6x - 4x)$
$= \sin(10x) \sin(2x)$
$= \sin 2x \sin 10x$
$= R.H.S.$
Explore More
Similar Questions
$\tan 15^{\circ}$ ની કિંમત શોધો:
Medium
View Solutionજો $A + B = \frac{\pi}{4}$ હોય,તો $(1 + \tan A)(1 + \tan B) = $
Medium
View Solutionજો $\frac{1-\tan \theta}{1+\tan \theta}=\frac{1}{\sqrt{3}}$,જ્યાં $\theta \in \left(0, \frac{\pi}{2}\right)$,તો $\theta=$
જો $\sin \alpha = 1/\sqrt{5}$ અને $\sin \beta = 3/5$ હોય,તો $\beta - \alpha$ કયા અંતરાલમાં આવે છે?
Difficult
View Solutionજો $\tan \theta = \frac{\cos 25^{\circ} + \sin 25^{\circ}}{\cos 25^{\circ} - \sin 25^{\circ}}$ અને $\theta$ ત્રીજા ચરણમાં હોય,તો $\theta =$ ($^{\circ}$ માં)
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