JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Trigonometric Ratios and Identities

Since the circumcentre is at the origin, the vertices lie at distance $R$ from it and may be written $A(R\cos \alpha ,R\sin \alpha )$, $B(R\cos \beta ,R\sin \beta )$ and $C(R\cos \gamma ,R\sin \gamma )$, so that the slopes of $OA$, $OB$ and $OC$ are $\tan \alpha$, $\tan \beta$ and $\tan \gamma$. For a triangle whose circumcentre is the origin the orthocentre is the vector sum $A+B+C$, and its lying on the $y$-axis makes the sum of the $x$-coordinates vanish, so $\cos \alpha +\cos \beta +\cos \gamma =0$. Writing $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$, the numerator becomes $4\left({\cos }^3\alpha +{\cos }^3\beta +{\cos }^3\gamma \right)-3\left(\cos \alpha +\cos \beta +\cos \gamma \right)$, and the second group is zero. Because $\cos \alpha +\cos \beta +\cos \gamma =0$, the identity $p^3+q^3+r^3=3pqr$ valid when $p+q+r=0$ gives ${\cos }^3\alpha +{\cos }^3\beta +{\cos }^3\gamma =3\cos \alpha \cos \beta \cos \gamma$. The numerator is therefore $12\cos \alpha \cos \beta \cos \gamma$, and dividing by $\cos \alpha \cos \beta \cos \gamma$ leaves $12$.