Question

If $\tan \alpha =\frac{1}{\sqrt{x(x^2+x+1)}},\tan \beta =\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$
and $\tan \gamma =\sqrt{x^{-3}+x^{-2}+x^{-1}}$, where $x\ne 0$, then $\alpha +\beta$ is

• A. $\gamma$
• B. $2\gamma$
• C. $-\gamma$
• D. $-2\gamma$

Answer 1

The correct option is A $\gamma$

We have,
$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{\frac{1}{\sqrt{x(x^2+x+1)}}+\frac{\sqrt{x}}{\sqrt{(x^2+x+1)}}}{1-\frac{1}{\sqrt{x(x^2+x+1)}}\frac{\sqrt{x}}{\sqrt{(x^2+x+1)}}}=\frac{(1+x)\sqrt{x^2+x+1}}{\sqrt{x}\cdot x(x+1)}=\sqrt{x^{-3}+x^{-2}+x^{-1}}=\tan \gamma$
$\therefore \alpha +\beta =\gamma$