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

A
A.P
B
G.P.
C
H.P
D
none of these

Solution:

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

If A, B, C are the angles of a triangle such that $\cot \frac{A}{2} = 3 \tan \frac{C}{2}$ , then sin A, sin B, sin C are in

Solution:

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