Question

In $\Delta ABC$, the least value of $cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}$ is

• A. $4\sqrt{2}$
• B. $0$
• C. $\frac{3}{\sqrt{2}}$
• D. $6$

Answer 1

The correct option is D $6$

For a triangle, the maximum or minimum occurs when the triangle is equilateral.

Therefore, $\angle A=\angle B=\angle C={60}^o$

$cosec\left(\frac{A}{2}\right)=cosec\left(\frac{B}{2}\right)=cosec\left(\frac{C}{2}\right)=\frac{1}{sin\frac{60}{2}}=\frac{1}{sin30}=2$

we know that $A.M.\ge G.M.$

$⟹\frac{cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}}{3}\ge \sqrt[3]{3cosec\frac{A}{2}\ast cosec\frac{B}{2}\ast cosec\frac{C}{2}}$

$⟹\frac{cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}}{3}\ge \sqrt[3]{32\ast 2\ast 2}$

$⟹cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}\ge 3\ast \sqrt[3]{32^3}$

$⟹cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}\ge 3\ast 2$

$⟹cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}\ge 6$

Therefore, the least value for $cosec\frac{A}{2}+cosec\frac{B}{2}+cosec\frac{C}{2}=6$