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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Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Question : If $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
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