Prove that: 1/[sin(x−a)sin(x−b)] = [cot(x−a) − cot(x−b)]/sin(a−b)

Question

Prove that:

\[ \frac{1}{\sin(x-a)\sin(x-b)} = \frac{\cot(x-a)-\cot(x-b)} {\sin(a-b)} \]

Proof

R.H.S.

\[ = \frac{\cot(x-a)-\cot(x-b)} {\sin(a-b)} \]

\[ = \frac{ \frac{\cos(x-a)}{\sin(x-a)} – \frac{\cos(x-b)}{\sin(x-b)} } {\sin(a-b)} \]

\[ = \frac{ \cos(x-a)\sin(x-b) – \sin(x-a)\cos(x-b) } {\sin(a-b)\sin(x-a)\sin(x-b)} \]

Using

\[ \sin C\cos D-\cos C\sin D = \sin(C-D) \]

\[ = \frac{ \sin[(x-b)-(x-a)] } {\sin(a-b)\sin(x-a)\sin(x-b)} \]

\[ = \frac{\sin(a-b)} {\sin(a-b)\sin(x-a)\sin(x-b)} \]

\[ = \frac{1} {\sin(x-a)\sin(x-b)} \]

L.H.S. = R.H.S.

Hence proved.