Trigonometric Identities Question 356
Question: If $ y=(1+\tan A)(1-\tan B) $ where $ A-B=\frac{\pi }{4} $ , then $ {{(y+1)}^{y+1}} $ is equal to
[J & K 2005]
Options:
A) 9
B) 4
C) 27
D) 81
Show Answer
Answer:
Correct Answer: C
Solution:
$ A-B=\frac{\pi }{4},\Rightarrow ,\tan ,(A-B)=\tan \frac{\pi }{4} $
$ \Rightarrow \frac{\tan A-\tan B}{1+\tan A,\tan B}=1 $
$ \Rightarrow \tan A-\tan B-\tan A,\tan B=1 $
$ \Rightarrow \tan A-\tan B-\tan A,\tan B+1=2 $
$ \Rightarrow (1+,\tan A)(1-\tan B)=2 $
Þ $ y=2 $ Hence, $ {{(y+1)}^{y+1}}={{(2+1)}^{2+1}}={{(3)}^{3}}=27 $ . Trick : Put suitable A and B as $ A-B=\frac{\pi }{4} $ i.e., $ A=\frac{\pi }{4},B=0 $
$ \therefore ,( 1+\tan \frac{\pi }{4} ),(1-\tan 0^{o})=2(1)=2 $ .
BETA
