Question

If A, B, C are in A.P., then \[\frac{\sin A - \sin C}{\cos C - \cos A}\]=

 

विकल्प

•  tan B

• cot B

• tan 2 B

• None of these

Answer 1

 cot B
Since A,B and C are in A.P,
B - A = C - B
or, 2B = A + C
\[\frac{\sin A - \sin C}{\cos C - \cos A}\]
\[ = \frac{2\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{- 2\sin\left( \frac{C + A}{2} \right)\sin\left( \frac{C - A}{2} \right)} \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \text{ and }\cos A - \cos B = - 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ = \frac{\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{- \sin\left( \frac{A + C}{2} \right)\sin\left( \frac{C - A}{2} \right)}\]
\[= \frac{\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{\sin\left( \frac{A + C}{2} \right)\sin\left( \frac{A - C}{2} \right)}\]
\[ = \frac{\cos\left( \frac{A + C}{2} \right)}{\sin\left( \frac{A + C}{2} \right)}\]
\[ = \frac{\cos B}{\sin B}\]
\[ = \cot B\]