Question

The least value of $\sec A+\sec B+\sec C$ in an acute angled triangle is

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

Answer 1

The correct option is D $6$

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

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

$secA=secB=secC=\frac{1}{cos60}=2$

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

$⟹\frac{secA+secB+secC}{3}\ge \sqrt[3]{3secA\ast secB\ast secC}$

$⟹\frac{secA+secB+secC}{3}\ge \sqrt[3]{32\ast 2\ast 2}$

$⟹secA+secB+secC\ge 3\ast \sqrt[3]{32^3}$

$⟹secA+secB+secC\ge 3\ast 2$

$⟹secA+secB+secC\ge 6$

Therefore, the least value for $secA+secB+secC=6$