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)

See https://math.stackexchange.com/questions/446402/if-alpha-beta-dfrac-pi4-prove-that-1-tan-alpha1-tan-beta and https://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt — lab bhattacharjee, Feb 12 '19 at 09:50

score 3 · Accepted Answer · edited Feb 11 '19 at 05:01

3

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 $(y+1)^{y+1}=3^3=27$.

edited Feb 11 '19 at 05:01

SNEHIL SANYAL

1,036

answered Feb 11 '19 at 03:19

GReyes

16,446
11
16

2

If y = 2 shouldn't y+1=3 and $(y+1)^{y+1}$ = $3^{3}$ = 27 – Feb 11 '19 at 03:27

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

1 Answers1