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Prove that:

Question:

Prove that:

(i) $\frac{1}{\sin (x-a) \sin (x-b)}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$

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

 

(iii) $\frac{1}{\cos (x-a) \cos (a-b)}=\frac{\tan (x-b)-\tan (x-a)}{\sin (a-b)}$

Solution:

(i) $\mathrm{RHS}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$

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

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

$=\frac{\sin (x-b-x+a)}{\sin (x-a) \sin (x-b) \sin (a-b)}$

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

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

= LHS

Hence proved.

(ii) $\mathrm{RHS}=\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)}$

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

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

$=\frac{\cos (x-b-x+a)}{\cos (a-b) \sin (x-a) \cos (x-b)} \quad($ Using $\quad \cos (A-B)=\cos A \cos b B+\sin A \sin B)$

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

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

= RHS

Hence proved.

(iii) RHS $=\frac{\tan (x-b)-\tan (x-a)}{\sin (a-b)}$

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

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

$=\frac{\sin (x-b-x+a)}{\sin (a-b) \cos (x-a) \cos (x-b)} \quad($ Using $\sin (A-B)=\sin A \cos B-\cos A \sin B)$

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

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

= LHS

Hence proved.

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