Question

If $\tan \alpha =\frac{m}{m-1}$ , $\tan \beta =\frac{1}{2m-1},then\alpha -\beta =$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\pi$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$
$=\frac{\frac{m}{m-1}-\frac{1}{2m-1}}{1+\frac{m}{(m-1)(2m-1)}}$ Given $\tan \alpha =\frac{m}{m-1},\tan \beta =\frac{1}{2m-1}$
$=\frac{2m^2-m-m+1}{2m^2-m-2m+1+m}$
$\tan (\alpha -\beta )=\frac{2m^2-2m+1}{2m^2-2m+1}=1$
$\Rightarrow \alpha -\beta =\frac{\pi }{4}$