TS EAMCET · Maths · Properties of Triangles
If \(A+C=2 B\), then \(\frac{\cos C-\cos A}{\sin A-\sin C}\) is equal to
- A \(\cot B\)
- B \(\cot 2 B\)
- C \(\tan 2 B\)
- D \(\tan B\)
Answer & Solution
Correct Answer
(B) \(\cot 2 B\)
Step-by-step Solution
Detailed explanation
Given that A + C += 2B ...(i) \(\begin{aligned} & \text { Now, } \frac{\cos C-\cos A}{\sin A-\sin C} \\ &= \frac{2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}{2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}\end{aligned}\)…
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