Question

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

Answer 1

$\tan \left(\alpha \right)=\frac{1}{\sqrt{x(x^2+x+1)}}\tan \left(\beta \right)=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\tan \left(\gamma \right)=\sqrt{x^{-3}+x^{-2}+x^{-1}}\alpha +\beta =\gamma \therefore \tan (\alpha +\beta )=\tan \left(\gamma \right)LHS=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}=\frac{\frac{1}{\sqrt{x(x^2+x+1)}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1-\frac{1}{x^2+x+1}}=\frac{x+1}{\sqrt{x(x^2+x+1)}}\times \frac{x^2+x+1}{x^2+x}=\frac{1}{\sqrt{x}}\times \frac{\sqrt{x^2+x+1}}{x}=\sqrt{\frac{x+1+x^{-1}}{x^3}}=\sqrt{{x-}^{-1}+x^{-2}+x^{-3}}=\left(\gamma \right)$$\alpha +\beta =\gamma =RHS$