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]

• A. 9
• B. 4
• C. 27
• D. 81

Answer 1

The correct option is C

27

A - B = $\frac{\pi }{4}$ $\Rightarrow$ tan (A - B) = tan $\frac{\pi }{4}$

$\Rightarrow$ $\frac{tanA-tanB}{1+tanAtanB}$ = 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 $\Rightarrow$ 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 ∴ (1 + tan$\frac{\pi }{4}$) (1 - $tan{0}^0$) = 2(1) = 2.