MHT CET · Maths · Trigonometric Ratios & Identities
In \((0,2 \pi)\), the number of solutions of \(\tan \theta+\sec \theta=2 \cos \theta\) are
- A 0
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\(\tan \theta+\sec \theta=2 \cos \theta \)
\( \Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=2 \cos \theta \)
\( \Rightarrow \sin \theta+1=2 \cos ^2 \theta \)
\( \Rightarrow \sin \theta+1=2\left(1-\sin ^2 \theta\right) \)
\( \Rightarrow 2 \sin ^2 \theta+\sin \theta-1=0\)
\(\therefore \sin \theta= -1, \frac{1}{2} \)
\(\text { For } \sin \theta =-1, \sin \theta=\frac{1}{2} \)
\(\theta =\frac{3 \pi}{2} \quad \theta=\frac{\pi}{6}, \frac{5 \pi}{6}\)
\(\therefore\) Number of solutions \(=3\)
\( \Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=2 \cos \theta \)
\( \Rightarrow \sin \theta+1=2 \cos ^2 \theta \)
\( \Rightarrow \sin \theta+1=2\left(1-\sin ^2 \theta\right) \)
\( \Rightarrow 2 \sin ^2 \theta+\sin \theta-1=0\)
\(\therefore \sin \theta= -1, \frac{1}{2} \)
\(\text { For } \sin \theta =-1, \sin \theta=\frac{1}{2} \)
\(\theta =\frac{3 \pi}{2} \quad \theta=\frac{\pi}{6}, \frac{5 \pi}{6}\)
\(\therefore\) Number of solutions \(=3\)
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