Question

prove that sin^2 6x-sin^2 4X=sin 2X. sin 10X ?

Answer 1

$\bold{\huge{\underline{Solution-}}}$

$\bf \: Used \: formula \\ \boxed{ \bf 1. {cos}^{2} A - {cos}^{2} B = sin(B + A) \: sin(B - A)} \\ \boxed{ \bf \: 2. \: {sin}^{2} A \: = 1 - {cos}^{2}A }$

$\bf \: L.H.S \\ \bf \:{sin}^{2}6x - {sin}^{2} 4x \\ \\ \implies \bf \: (1 - {cos}^{2} 6x) - (1 - {cos}^{2} 4x) \: \\ \\ \implies \bf \: {cos}^{2} 4x - {cos}^{2} 6x \\ \\ \implies \bf \: sin(6x + 4x) \: sin(6x - 4x) \\ \\ \text{ \: now \: using \: the \: formula \:} \bf \: {cos}^{2} A - {cos}^{2} B \\ \\ \implies \bf \: sin \: 10x \: sin \: 2x \\ \\ \implies \bf \: sin \: 2x \: sin \: 10x$