Prove that: 1/[cos(x – a) cos(x – b)] = [tan(x – b) – tan(x – a)] / sin(a – b)
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)]
Prove that: 1/[cos(x – a) cos(x – b)] = [tan(x – b) – tan(x – a)] / sin(a – b) Read More »