Question

Let angle A, B and C of a triangle ABC be in arithmetic progression. If $\frac{b}{c}=\sqrt{\frac{3}{2}}$ then the value of cosec 2A is equal to
(Symbols used have usual meaning in a triangle ABC)

Answer 1

Since $A,B,C$ are in AP, then B = $60$.
Also, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=>\frac{b}{c}=\frac{\sin B}{\sin C}=>\frac{\frac{\sqrt{3}}{2}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}=>\sin C=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=>C={45}^o=>A=180-60-45={75}^o=>\sin \left(2A\right)=\sin {150}^o=0.5=>cosec\left(2A\right)=\frac{1}{0.5}=2$
Hence, answer is $2$.