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
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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