Question

If $y=(1+\tan A)(1-\tan B),$ where $A-B=\frac{\pi }{4},$ then ${(y+1)}^{y+1}$ is equal to

• A. $9$
• B. $4$
• C. $27$
• D. $81$

Answer 1

Answer: (C)

Given, $A-B=\frac{\pi }{4}$
$\Rightarrow$ $\tan (A-B)=\tan \frac{\pi }{4}$
$\Rightarrow$ $\frac{\tan A-tanB}{1+\tan A\tan B^{-1}}$ $\Rightarrow$
$\tan A-\tan B-\tan A\tan B=1$ ..(i)
Also given, $y=(1+\tan A)(1-\tan B)$
$=(1-tanB+\tan A-\tan A\tan B)$
$=(1+1)$ [from Eq. (i)] $=2$
$\therefore$ ${(y+1)}^{y+1}={(2+1)}^{2+1}=3^3=27$