Question

If $$\tan \alpha =\frac{x}{x+1}$$ and 

$$\tan \beta =\frac{1}{2x+1}$$, then

$$\tan \beta =\frac{1}{2x+1}$$ is equal to

 

विकल्प

• $$\frac{\pi }{2}$$

   

• $$\frac{\pi }{2}$$

   

• $$\frac{\pi }{2}$$

   

• $$\frac{\pi }{2}$$

   

Answer 1

It is given that $$\tan \alpha =\frac{x}{x+1}$$ and $$\tan \beta =\frac{x}{x+1}$$ 

Now,
$$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$
$$=\frac{\frac{x}{x+1}+\frac{1}{2x+1}}{1-\frac{x}{x+1}\times \frac{1}{2x+1}}$$
$$=\frac{\frac{x(2x+1)+x+1}{(x+1)(2x+1)}}{\frac{(x+1)(2x+1)-x}{(x+1)(2x+1)}}$$
$$=\frac{2x^2+x+x+1}{2x^2+3x+1-x}$$
$$=\frac{2x^2+2x+1}{2x^2+2x+1}$$
$$=1$$
$$\therefore \tan (\alpha +\beta )=1=\tan \frac{\pi }{4}$$
$$\Rightarrow \alpha +\beta =\frac{\pi }{4}$$

Hence, the correct answer is option D.