JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 4)

If $$\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$$ is the solution of $$4 \cos \theta+5 \sin \theta=1$$, then the value of $$\tan \alpha$$ is

$$\frac{10-\sqrt{10}}{12}$$

$$\frac{\sqrt{10}-10}{6}$$

$$\frac{\sqrt{10}-10}{12}$$

$$\frac{10-\sqrt{10}}{6}$$

$$4+5 \tan \theta=\sec \theta$$

Squaring : $$24 \tan ^2 \theta+40 \tan \theta+15=0$$

$$\tan \theta=\frac{-10 \pm \sqrt{10}}{12}$$

and $$\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right)$$ is Rejected.

(3) is correct.