Prove the following

Question:

Let $\tan \alpha, \tan \beta$ and $\tan \gamma ; \alpha, \beta, \gamma \neq \frac{(2 n-1) \pi}{2}$

$\mathrm{n} \in \mathrm{N}$ be the slopes of three line segments $\mathrm{OA} \mathrm{OB}$ and $\mathrm{OC}$, respectively, where $\mathrm{O}$ is origin.If circumcentre of $\Delta \mathrm{ABC}$ coincides with origin and its orthocentre lies on $y$-axis, then the value of $\left(\frac{\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma}{\cos \alpha \cos \beta \cos \gamma}\right)^{2}$ is equal to :

Solution:

Since orthocentre and circumcentre both lies on $y$-axis

$\Rightarrow$ Centroid also lies on $y$-axis

$\Rightarrow \Sigma \cos \alpha=0$

$\cos \alpha+\cos \beta+\cos \gamma=0$

$\Rightarrow \cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma=3 \cos \alpha \cos \beta \cos \gamma$

$\therefore \frac{\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma}{\cos \alpha \cos \beta \cos \gamma}$

$=\frac{4\left(\cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma\right)-3(\cos \alpha+\cos \beta+\cos \gamma)}{\cos \alpha \cos \beta \cos \gamma}$

$=12$

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