Question

If A + B = $\frac{\pi }{4}$ where A, B $\in R^{+}$, then the minimum value of (1+tanA) (1+tanB) is always equal to

• A. 2
• B. 4
• C. 1
• D. 0

Answer 1

The correct option is A

2

A + B = $\frac{\pi }{4}\Rightarrow tan(A+B)=1$

$\Rightarrow$ tanA + tanB = 1 - tanA. tanB

$\Rightarrow$ tanA + tanB + tanA tanB + 1 = 2

$\Rightarrow$ (1 + tanA)(1 + tanB) = 2