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

Question

Prove that:

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

Proof

R.H.S.

\[ = \frac{\cot(x-a)+\tan(x-b)} {\cos(a-b)} \]

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

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

Using

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

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

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

Since

\[ \cos(b-a)=\cos(a-b) \]

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

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

Hence proved.