Question

If $\tan \alpha =\frac{1}{\sqrt{x(x^2+x+1)}},\tan \beta =\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan \gamma =\frac{\sqrt{x^2+x+1}}{x\sqrt{x}}$ show that $\alpha +\beta =\gamma$.

Answer 1

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

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

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

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

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

$=\tan \gamma$

$\Rightarrow \alpha +\beta =\gamma$