If $y = (1 + \tan A)(1 - \tan B)$, where $A - B = \frac{\pi}{4}$ , then find the value of $(y+1)^{y+1}$.

Question

If $y = (1 + \tan A)(1 - \tan B)$, where $A - B = \frac{\pi}{4}$ , then find the value of $(y+1)^{y+1}$.

I have tried this question and I think that it is something along the lines of $\tan (A-B) = \frac{\tan A - tan B}{1 + \tan A \tan B} \Rightarrow 1- \tan A \tan B = \tan A \tan B$.

Also if you solved this question, can you rate the difficulty level of this question (I just want to understand where I stand)

Answer 1

You have $$ 1=\tan (\pi/4)=\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} $$ which implies $$ \tan A-\tan B=1+\tan A\tan B $$ Your quantity is then $$ y=(1+\tan A)(1-\tan B)=1+\tan A-\tan B-\tan A\tan B=2 $$ hence $(y+1)^{y+1}=3^3=27$.