Question

If A + B = π/3where A, B > 0 then minimum value of

secA + secB is equal to

• A. 4
• B. 8
• C. 6
• D. None of these

Answer 1

Answer: (A)

4

Answer 2

Clearly A, B ∈ $\left( 0 , \frac { \pi } { 3 } \right)$. For y = sec x, $\frac { d y } { d x }$ and $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ are positive in $\left( 0 , \frac { \pi } { 3 } \right)$. Hence

sec A + sec B ≥ 2 sec$\left( \frac { A + B } { 2 } \right)$ = $2 \sec \frac { \pi } { 6 } = \frac { 4 } { \sqrt { 3 } }$