# If

Question:

If $\alpha+\beta=\frac{\pi}{4}$, then the value of $(1+\tan \alpha)(1+\tan \beta)$ is

(a) 1

(b) 2

(c) –2

(d) not defined

Solution:

Given $\alpha+\beta=\frac{\pi}{4}$

$(1+\tan \alpha)(1+\tan \beta)$

$1+\tan \alpha+\tan \beta+\tan \alpha \tan \beta$

$=1+\tan (\alpha+\beta)(1-\tan \alpha \tan \beta)+\tan \alpha \tan \beta$

using identity : $\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a \tan b}$

$=1+\tan (\pi / 4)(1-\tan \alpha \tan \beta)+\tan \alpha \tan \beta(\because \alpha+\beta=\pi / 4$ given $)$

$=1+1(1-\tan \alpha \tan \beta)+\tan \alpha \tan \beta$

$=1+1-\tan \alpha \tan \beta+\tan \alpha \tan \beta$

$=2$

Hence, $(1+\tan \alpha)(1+\tan \beta)=2$

Hence, the corrrect answer is option B.

Since $\alpha+\beta=\pi / 4$

$\Rightarrow \tan (\alpha+\beta)=\tan \pi / 4$

$\Rightarrow \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=1 \quad$ (using identity : $\left.\tan (a+b)=\frac{\tan a+\tan b}{1-\tan \tan b}\right)$

$\Rightarrow \tan \alpha+\tan \beta=1-\tan \alpha \tan \beta$

$\Rightarrow \tan \alpha+\tan \alpha \tan \beta+\tan \beta=1$

$\Rightarrow 1+\tan \alpha+\tan \beta(1+\tan \alpha)=1+1$

$\Rightarrow(1+\tan \alpha)(1+\tan \beta)=2$

Hence, the correct answer option B.