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 $