Question

If $\tan \alpha =\frac{x}{x+1},\tan \beta =\frac{1}{2x+1}$, then value of $(\alpha +\beta )$ may be :

• A. $\frac{\pi }{6}$
• B. $\frac{13\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. $\frac{11\pi }{4}$

Answer 1

The correct option is B $\frac{13\pi }{4}$

$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$=\frac{\frac{x}{x+1}+\frac{1}{2x+1}}{1-\frac{x}{x+1}\times \frac{1}{2x+1}}$

$=\frac{\frac{2x^2+x+x+1}{(x+1)(2x+1)}}{\frac{(x+1)(2x+1)-x}{(x+1)(2x+1)}}$

$=\frac{2x^2+2x+1}{2x^2+x+2x+1-x}$

$=\frac{2x^2+2x+1}{2x^2+2x+1}$

$=1$

$=\tan \frac{\pi }{4}$

$\therefore \alpha +\beta =\frac{\pi }{4},\pi +\frac{\pi }{4},2\pi +\frac{\pi }{4},3\pi +\frac{\pi }{4},..$

$=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},\frac{13\pi }{4},..$