Question

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

• A. $2$
• B. $4$
• C. $1$
• D. $Noneofthese$

Answer 1

The correct option is A $2$

As it is given that $A+B=\frac{\pi }{4}$, then,

$\tan \left(A+B\right)=\tan \frac{\pi }{4}$

$\frac{\tan A+\tan B}{1-\tan A\tan B}=1$

$\tan A+\tan B=1-\tan A\tan B$ (1)

Simplify $\left(1+\tan A\right)\left(1+\tan B\right)$,

$\left(1+\tan A\right)\left(1+\tan B\right)=1+\tan B+\tan A+\tan A\tan B$

Then from equation (1),

$\left(1+\tan A\right)\left(1+\tan B\right)=1+1-\tan A\tan B+\tan A\tan B$

$\left(1+\tan A\right)\left(1+\tan B\right)=2$