JEE Advanced 2017 · Paper 2, Mathematics question 9

Official key: marks to all.

Solve for cos β. Rearranging,
$$(1+2\cos\alpha)=\cos\beta\,(2+\cos\alpha)\implies\cos\beta=\frac{1+2\cos\alpha}{2+\cos\alpha}.$$

Half-angle form. Put $a=\tan^2\frac\alpha2$, so $\cos\alpha=\frac{1-a}{1+a}$. Then
$$\cos\beta=\frac{3-a}{3+a}=\frac{1-a/3}{1+a/3}\implies\tan^2\frac\beta2=\frac13\tan^2\frac\alpha2.$$

Conclusion. So $\tan\frac\alpha2=\pm\sqrt3\tan\frac\beta2$. Either (A) or (C) holds, depending on the signs, but neither is true in every case. The question was given marks to all.