# If A, B, C Are in A.P., Then Sin a − Sin C Cos C − Cos a = - Mathematics

MCQ
Sum

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

#### Options

•  tan B

• cot B

• tan 2 B

• None of these

#### Solution

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

Concept: Transformation Formulae
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Q 11 | Page 21