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

$\begin{array}{1 1}(A)\;1&(B)\;2\\(C)\;-2&(D)\;\text{Not defined}\end{array}$

Toolbox:
• $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
Given
$\alpha+\beta=\large\frac{\pi}{4}$
$\tan(\alpha+\beta)=\large\frac{\tan \alpha+\tan \beta}{1-\tan \alpha\tan \beta}$
$\tan\large\frac{\pi}{4}=\large\frac{\tan \alpha+\tan \beta}{1-\tan \alpha\tan \beta}$
$\therefore 1-\tan \alpha\tan \beta=\tan\alpha+\tan\beta$-------(1)
$(1+\tan\alpha)(1+\tan \beta)=\tan\alpha+\tan\beta+1+\tan\alpha.\tan\beta$
From (1) we get
$1-\tan \alpha\tan \beta+1+\tan \alpha\tan \beta$
$\Rightarrow 2$
Hence (B) is the correct answer.
answered Apr 21, 2014