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Question: If \(y = (1 + \tan A)(1 - \tan B)\) where \(A - B = \frac{\pi}{4}\), then \((y + 1)^{y + 1}\) is equ...

If y=(1+tan⁡A)(1−tan⁡B)y = (1 + \tan A)(1 - \tan B) where A−B=π4A - B = \frac{\pi}{4}, then (y+1)y+1(y + 1)^{y + 1} is equal to

A

9

B

4

C

27

D

81

Answer

27

Explanation

Solution

A−B=π4⇒tan⁡(A−B)=tan⁡π4⇒tan⁡A−tan⁡B1+tan⁡Atan⁡B=1A - B = \frac{\pi}{4} \Rightarrow \tan(A - B) = \tan\frac{\pi}{4} \Rightarrow \frac{\tan A - \tan B}{1 + \tan A\tan B} = 1

⇒tan⁡A−tan⁡B−tan⁡Atan⁡B=1\Rightarrow \tan A - \tan B - \tan A\tan B = 1

⇒tan⁡A−tan⁡B−tan⁡Atan⁡B+1=2\Rightarrow \tan A - \tan B - \tan A\tan B + 1 = 2

⇒(1+tan⁡A)(1−tan⁡B)=2\Rightarrow (1 + \tan A)(1 - \tan B) = 2 ⇒ y=2y = 2

Hence, (y+1)y+1=(2+1)2+1=(3)3=27(y + 1)^{y + 1} = (2 + 1)^{2 + 1} = (3)^{3} = 27.

Trick : Put suitable A and B as A−B=π4A - B = \frac{\pi}{4}

i.e.,A=π4,B=0A = \frac{\pi}{4},B = 0 ∴(1+tan⁡π4)(1−tan⁡0o)=2(1)=2\therefore\left( 1 + \tan\frac{\pi}{4} \right)(1 - \tan 0^{o}) = 2(1) = 2.