If A, B, C are the angles of a triangle such that cot⁡A2=3 tan⁡C2, then sin A, sin B, sin C are in

# If A, B, C are the angles of a triangle such that , then sin A, sin B, sin C are in

1. A

A.P

2. B

G.P.

3. C

H.P

4. D

none of these

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### Solution:

Given $\mathrm{cot}\frac{\mathrm{A}}{2}\cdot \mathrm{cot}\frac{\mathrm{C}}{2}=3$
$⇒\frac{\mathrm{cos}\frac{\mathrm{A}}{2}\cdot \mathrm{cos}\frac{\mathrm{C}}{2}}{\mathrm{sin}\frac{\mathrm{A}}{2}\cdot \mathrm{sin}\frac{\mathrm{C}}{2}}=3$
$⇒\frac{\mathrm{cos}\frac{\mathrm{A}-\mathrm{C}}{2}}{\mathrm{cos}\frac{\mathrm{A}+\mathrm{C}}{2}}=2$ (using componendo and dividendo)
$\begin{array}{l}⇒\frac{2\mathrm{sin}\frac{\mathrm{A}+\mathrm{C}}{2}\mathrm{cos}\frac{\mathrm{A}-\mathrm{C}}{2}}{2\mathrm{sin}\frac{\mathrm{A}+\mathrm{C}}{2}\cdot \mathrm{cos}\frac{\mathrm{A}+\mathrm{C}}{2}}=2\\ ⇒2\mathrm{sin}\mathrm{B}=\mathrm{sin}\mathrm{A}+\mathrm{sin}\mathrm{C}\end{array}$  