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

Question

Prove that:

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

Proof

R.H.S.

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

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

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

Using

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

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

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

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

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

Hence proved.