Solve this trigonometric problem to find if angles A, B, C are in AP, GP, or HP based on (sin A - sin C)/(cos C - cos A) = cot B.

MHT- CET Trigonometric Problem

If \( \frac{\sin A - \sin C}{\cos C - \cos A} = \cot B \), then \( A \), \( B \), and \( C \) are in:

a) AP

b) GP

c) HP

d) none of the above

Detailed Solution

Problem: \( \frac{\sin A - \sin C}{\cos C - \cos A} = \cot B \). Find the relationship between \( A \), \( B \), and \( C \).

Step 1: We need to simplify the left side using trigonometric identities.

Step 2: Use Sum-to-Product Identities:

• \( \sin A - \sin C = 2 \cos \left( \frac{A + C}{2} \right) \sin \left( \frac{A - C}{2} \right) \)
• \( \cos C - \cos A = 2 \sin \left( \frac{A + C}{2} \right) \sin \left( \frac{A - C}{2} \right) \) (since \( \cos C - \cos A = -2 \sin \left( \frac{C + A}{2} \right) \sin \left( \frac{C - A}{2} \right) \), and adjusting signs)

Step 3: Substitute:

\( \frac{\sin A - \sin C}{\cos C - \cos A} = \frac{2 \cos \left( \frac{A + C}{2} \right) \sin \left( \frac{A - C}{2} \right)}{2 \sin \left( \frac{A + C}{2} \right) \sin \left( \frac{A - C}{2} \right)} = \frac{\cos \left( \frac{A + C}{2} \right)}{\sin \left( \frac{A + C}{2} \right)} = \cot \left( \frac{A + C}{2} \right) \)

Step 4: Equate:

\( \cot \left( \frac{A + C}{2} \right) = \cot B \)

Since \( \cot x = \cot y \) implies \( x = y + n\pi \), we get \( \frac{A + C}{2} = B + n\pi \). Assuming \( n = 0 \) (simplest case):

\( \frac{A + C}{2} = B \Rightarrow 2B = A + C \)

Step 5: Conclusion: \( 2B = A + C \) means \( A \), \( B \), \( C \) are in Arithmetic Progression (AP).

Answer: a) AP