Question

In a triangle ABC, let $\sqrt{\sin A} + \sqrt{\sin B} + \sqrt{\sin C} = \sqrt{12 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$, then the value of $(\frac{\sin^2 A}{\sin^2 B} + \frac{\sin^2 B}{\sin^2 C} + \frac{\sin^2 C}{\sin^2 A})$, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The given condition is $\sqrt{\sin A} + \sqrt{\sin B} + \sqrt{\sin C} = \sqrt{12 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$. For an equilateral triangle ($A=B=C=60^\circ$), LHS = $3\sqrt{\sin 60^\circ} = 3\sqrt{\frac{\sqrt{3}}{2}}$ and RHS = $\sqrt{12 \cos^3 30^\circ} = \sqrt{12(\frac{\sqrt{3}}{2})^3} = \sqrt{12 \frac{3\sqrt{3}}{8}} = \sqrt{\frac{9\sqrt{3}}{2}}$. Squaring both sides, $9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}$, which matches.

Squaring the given equation and using identities, we can show that the condition implies $A=B=C=60^\circ$. Since $A=B=C$, $\sin A = \sin B = \sin C$. Therefore, $\frac{\sin^2 A}{\sin^2 B} = 1$, $\frac{\sin^2 B}{\sin^2 C} = 1$, and $\frac{\sin^2 C}{\sin^2 A} = 1$. The value of the expression is $1 + 1 + 1 = 3$.