Browse Questions

# If $\tan\alpha=\large\frac{1}{\sqrt{x(x^2+x+1)}}$,$\tan\beta=\large\frac{\sqrt x}{\sqrt{x^2+x+1}}$ and $\tan\gamma=\sqrt{x^{-3}+x^{-2}+x^{-1}}$ then $\alpha+\beta$ is

$(a)\;\gamma\qquad(b)\;2\gamma\qquad(c_\;-\gamma\qquad(d)\;None\;of\;these$

We have
$\tan(\alpha+\beta)=\large\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$\Rightarrow \large\frac{\Large\frac{1}{\sqrt{x(x^2+x+1)}}+\frac{\sqrt x}{\sqrt{x^2+x+1}}}{1-\Large\frac{1}{\sqrt{x(x^2+x+1)}}.\frac{\sqrt x}{\sqrt{x^2+x+1}}}$
$\Rightarrow \large\frac{(1+x)\sqrt{x^2+x+1}}{\sqrt x.x(x+1)}$
$\Rightarrow \sqrt{x^{-3}+x^{-2}+x^{-1}}$
$\Rightarrow \tan\gamma$
$\alpha+\beta=\tan^{-1}\tan\gamma$
$\alpha+\beta=\gamma$
Hence (a) is the correct answer.