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
$\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}$ નું મૂલ્ય શું છે?
DifficultKCET 2017
View Solution$\cos 15^{\circ} - \sin 15^{\circ}$ ની કિંમત ...... છે.
DifficultMHT CET 2026
View Solutionજો $\cos \alpha + \cos \beta = \frac{1}{3}$ અને $\sin \alpha + \sin \beta = \frac{1}{4}$ હોય,તો $\cos (\alpha + \beta) = $
$\triangle ABC$ માં,જો $A = \tan^{-1} 2$ અને $B = \tan^{-1} 3$ હોય,તો $C = $ . . . . . . ($^{\circ}$ માં)
સાબિત કરો કે: $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$
Medium
View SolutionVedclass 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