Question

$\text{The smallest value of}\left(\theta \right)\text{satisfying the equation}\sqrt{3}(cot\theta +tan\theta )=4is$

• A. $2\pi /3$
• B. $\pi /3$
• C. $\pi /6$
• D. $\pi /12$

Answer 1

The correct option is C

$\pi /6$

$Given:\sqrt{3}(cot\theta +tan\theta )=4\Rightarrow \sqrt{3}(\frac{cos\theta }{sin\theta }+\frac{sin\theta }{cos\theta })\Rightarrow \sqrt{3}(co{s}^2\theta +si{n}^2\theta )=4sin\theta cos\theta \Rightarrow \sqrt{3}=2sin2\theta [sin2\theta =2sin\theta cos\theta ]\Rightarrow sin2\theta =\frac{\sqrt{3}}{2}\Rightarrow sin2\theta =sin\frac{\pi }{2}\Rightarrow 2\theta =n\pi +(-1{)}^n\frac{\pi }{3},n\in Z\Rightarrow \theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{6},n\in Z\text{To obrain the smallest value of}\theta ,wewillputn=0intheaboveequation.Thus,wehave,\theta =\frac{\pi }{6}\text{Hence,the smallest value of}\theta is\frac{\pi }{6}.$