Prove that: \[ \frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A} = \cot 8A \]
Solution
Consider the left-hand side:
\[
\frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A}
\]
Using the identity:
\[
\sin A-\sin B
=
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
\sin 9A-\sin 7A
=
2\cos\frac{9A+7A}{2}\sin\frac{9A-7A}{2}
\]
\[
=
2\cos8A\sin A
\]
Now using:
\[
\cos A-\cos B
=
-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
\cos 7A-\cos 9A
=
-2\sin\frac{7A+9A}{2}\sin\frac{7A-9A}{2}
\]
\[
=
-2\sin8A\sin(-A)
\]
Since,
\[
\sin(-A)=-\sin A
\]
\[
=
2\sin8A\sin A
\]
Therefore,
\[
\frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A}
=
\frac{2\cos8A\sin A}{2\sin8A\sin A}
\]
\[
=
\frac{\cos8A}{\sin8A}
\]
\[
=
\cot8A
\]
Hence,
\[
\boxed{
\frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A}
=
\cot8A
}
\]