QPaperGen

We use the identity:


\[
\sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B)
\]



Let \( A = 6x \), \( B = 4x \). Then:
\begin{align*}
\text{LHS} &= \sin^2(6x) - \sin^2(4x) \\
&= (\sin 6x + \sin 4x)(\sin 6x - \sin 4x)
\end{align*}

Now apply the sum and difference identities:


\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]




\[
\sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]



So:
\begin{align*}
\sin 6x + \sin 4x &= 2 \sin(5x) \cos(x) \\
\sin 6x - \sin 4x &= 2 \cos(5x) \sin(x)
\end{align*}

Therefore:
\begin{align*}
\text{LHS} &= (2 \sin 5x \cos x)(2 \cos 5x \sin x) \\
&= 4 \sin 5x \cos 5x \sin x \cos x
\end{align*}

Now use the identities:


\[
\sin(2x) = 2 \sin x \cos x, \quad \sin(10x) = 2 \sin 5x \cos 5x
\]



So:


\[
\text{LHS} = \sin(2x) \cdot \sin(10x) = \text{RHS}
\]





\[
\boxed{\sin^2(6x) - \sin^2(4x) = \sin(2x) \cdot \sin(10x)}
\]