Prove that: \[ \frac{\sin A+\sin 3A}{\cos A-\cos 3A} = \cot A \]
Solution
Consider the left-hand side:
\[
\frac{\sin A+\sin 3A}{\cos A-\cos 3A}
\]
Using the identity:
\[
\sin A+\sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
\sin A+\sin 3A
=
2\sin\frac{A+3A}{2}\cos\frac{A-3A}{2}
\]
\[
=
2\sin2A\cos(-A)
\]
Since,
\[
\cos(-A)=\cos A
\]
\[
=
2\sin2A\cos A
\]
Now using:
\[
\cos A-\cos B
=
-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
\cos A-\cos 3A
=
-2\sin\frac{A+3A}{2}\sin\frac{A-3A}{2}
\]
\[
=
-2\sin2A\sin(-A)
\]
Since,
\[
\sin(-A)=-\sin A
\]
\[
=
2\sin2A\sin A
\]
Therefore,
\[
\frac{\sin A+\sin 3A}{\cos A-\cos 3A}
=
\frac{2\sin2A\cos A}{2\sin2A\sin A}
\]
\[
=
\frac{\cos A}{\sin A}
\]
\[
=
\cot A
\]
Hence,
\[
\boxed{
\frac{\sin A+\sin 3A}{\cos A-\cos 3A}
=
\cot A
}
\]