Medium MCQ +4 / -1 PYQ · JEE Mains 2024

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. A $\frac{10-\sqrt{10}}{12}$
  2. B $\frac{\sqrt{10}-10}{6}$
  3. C $\frac{\sqrt{10}-10}{12}$ Correct answer
  4. D $\frac{10-\sqrt{10}}{6}$

Solution

<p>$4+5 \tan \theta=\sec \theta$</p> <p>Squaring : $24 \tan ^2 \theta+40 \tan \theta+15=0$</p> <p>$\tan \theta=\frac{-10 \pm \sqrt{10}}{12}$</p> <p>and $\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right)$ is Rejected.</p> <p>(3) is correct.</p>

About this question

Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities

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