JEE Mains · Maths · STD 11 - Trigonometrical equations
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
- A \(\frac{10-\sqrt{10}}{6}\)
- B \(\frac{10-\sqrt{10}}{12}\)
- C \(\frac{\sqrt{10}-10}{12}\)
- D \(\frac{\sqrt{10}-10}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{10}-10}{12}\)
Step-by-step Solution
Detailed explanation
\(4+5 \tan \theta=\sec \theta\) 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.
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