QPaperGen

We use the identities:


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


Left-hand side:
\begin{align*}
\cot 4x \cdot (\sin 5x + \sin 3x)
&= \cot 4x \cdot \left[ 2 \sin\left( \frac{5x + 3x}{2} \right) \cos\left( \frac{5x - 3x}{2} \right) \right] \\
&= \cot 4x \cdot \left[ 2 \sin(4x) \cos(x) \right] \\
&= \frac{\cos 4x}{\sin 4x} \cdot 2 \sin(4x) \cos(x) \\
&= 2 \cos 4x \cos x
\end{align*}

Right-hand side:
\begin{align*}
\cot x \cdot (\sin 5x - \sin 3x)
&= \cot x \cdot \left[ 2 \cos\left( \frac{5x + 3x}{2} \right) \sin\left( \frac{5x - 3x}{2} \right) \right] \\
&= \cot x \cdot \left[ 2 \cos(4x) \sin(x) \right] \\
&= \frac{\cos x}{\sin x} \cdot 2 \cos(4x) \sin(x) \\
&= 2 \cos 4x \cos x
\end{align*}

Therefore:


\[
\text{LHS} = \text{RHS}
\quad \Rightarrow \quad
\boxed{\cot 4x \cdot (\sin 5x + \sin 3x) = \cot x \cdot (\sin 5x - \sin 3x)} \qquad \textbf{Hence proved.}
\]