Trigonometric Equations Question 4
Question 4 - 2024 (29 Jan Shift 1)
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
(1) $\frac{10-\sqrt{10}}{6}$
(2) $\frac{10-\sqrt{10}}{12}$
(3) $\frac{\sqrt{10}-10}{12}$
(4) $\frac{\sqrt{10}-10}{6}$
Show Answer
Answer (3)
Solution
$4+5 \tan \theta=\sec \theta$
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.
