Question

If $A+B={45}^{\circ }$ then $\left(1+\tan A\right)\left(1+\tan B\right)=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (D)

$2$

Given

$A+B=45$

$\tan (A+B)=\tan 45$

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

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

$\tan A+\tan B+\tan A\tan B=1$

Adding ${}^{\prime }{1}^{\prime }$ on both sides

$1+\tan A+\tan B+\tan A\tan B=2$

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

$(1+\tan A)(1+\tan B)=2$.