Question

In an acute angled triangle $ABC$, the least value of $\sec A+\sec B+\sec C$ is?

• A. $6$
• B. $3$
• C. $9$
• D. $4$

Answer 1

Answer: (A)

$6$

$cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow secA=\frac{2bc}{b^2+c^2-a^2}$

$secA$ is MIninmum$(=2)$ when $a=b=c=1$

Similarly Minimum Value for $secB=secC=2$

Therefore, Their Least Sum is $2+2+2=6$

Therefore, Correct Answer is $A$