MHT CET · Maths · Trigonometric Ratios & Identities
In a triangle \(\mathrm{ABC}\) if \(\frac{\sin \mathrm{A}-\sin \mathrm{C}}{\cos \mathrm{C}-\cos \mathrm{A}}=\cot \mathrm{B}\), then \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in
- A Arithmetico - Geometric progression
- B Harmonic Progression
- C Geometric progression
- D Arithmetic progression
Answer & Solution
Correct Answer
(D) Arithmetic progression
Step-by-step Solution
Detailed explanation
\(\frac{\sin A-\sin C}{\cos C-\cos A}=\cot B\)
\(2 \cos \left(\frac{A+C}{2}\right) \cdot \sin \left(\frac{A-C}{2}\right)\)
\(2 \cdot \sin \left(\frac{A+C}{2}\right) \cdot \sin \left(\frac{A-C}{2}\right)\)
\(\cot \left(\frac{A+C}{2}\right)=\cot B \Rightarrow \frac{A+C}{2}=B \Rightarrow A+C=2 B\)
\(\therefore A, B, C\) are in A.P.
\(2 \cos \left(\frac{A+C}{2}\right) \cdot \sin \left(\frac{A-C}{2}\right)\)
\(2 \cdot \sin \left(\frac{A+C}{2}\right) \cdot \sin \left(\frac{A-C}{2}\right)\)
\(\cot \left(\frac{A+C}{2}\right)=\cot B \Rightarrow \frac{A+C}{2}=B \Rightarrow A+C=2 B\)
\(\therefore A, B, C\) are in A.P.
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