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यदि समीकरण $4 \cos \theta+5 \sin \theta=1$. का हल $\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ है, तो $\tan \alpha$ का मान है
A
$\frac{10-\sqrt{10}}{6}$
B
$\frac{10-\sqrt{10}}{12}$
C
$\frac{\sqrt{10}-10}{12}$
D
$\frac{\sqrt{10}-10}{6}$
(JEE MAIN-2024)
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.
Standard 11
Mathematics
