Question : If $3+\cos ^2 \theta=3\left(\cot ^2 \theta+\sin ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\cos \theta+2 \sin \theta)$ ?
Option 1: $\frac{2 \sqrt{3}+1}{2}$
Option 2: $3 \sqrt{2}$
Option 3: $\frac{3 \sqrt{3}+1}{2}$
Option 4: $\frac{\sqrt{3}+2}{2}$
Correct Answer: $\frac{2 \sqrt{3}+1}{2}$
Solution :
According to the question
$3+\cos ^2 \theta=3\left(\cot ^2 \theta+\sin ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$,
⇒ $3 + \cos^{2}\theta$ = 3($\cot^{2}\theta + \sin^{2}\theta$)
⇒ $3 + \cos^{2}\theta$ = 3 $\cot^{2}\theta + 3(1 - \cos^{2}\theta$)
⇒ $3 + \cos^{2}\theta$ = 3 $\cot^{2}\theta + 3 - 3\cos^{2}\theta$
⇒ $4 \cos^{2}θ = 3(\frac {\cos^{2}θ} {\sin^{2}θ})$
⇒ $\sin^{2}θ$ = $\frac{3}{4}$
⇒ $\sin θ = \frac{\sqrt{3}}{2} = \sin 60°$
⇒ $θ = 60°$
Now,
$(\cos θ + 2 \sinθ) = (\cos 60° + 2 \sin 60°)
= (\frac{1}{2}) + 2(\frac{\sqrt{3}}{2})
= \frac{1}{2} + \sqrt{3}
= \frac{2\sqrt{3} + 1}{2}$
Hence, the correct answer is $\frac{2\sqrt{3} + 1}{2}$.
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