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 $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
Option 1: $1+2 \sqrt{3}$
Option 2: $\frac{1+2 \sqrt{3}}{3}$
Option 3: $\frac{2+\sqrt{3}}{3}$
Option 4: $2+\sqrt{3}$
Correct Answer: $2+\sqrt{3}$
Solution :
Given,
$\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4$
⇒ $\frac{\cos\theta(1+\sin\theta)+\cos\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}=4$
⇒ $\frac{\cos\theta+\sin\theta\cos\theta+\cos\theta-\sin\theta\cos\theta}{(1+\sin\theta)(1-\sin\theta)}=4$
⇒ $\frac{2\cos\theta}{1-\sin^2\theta}=4$ [As $\sin^2\theta+\cos^2\theta=1$]
⇒ $\frac{\cos\theta}{\cos^2\theta}=2$
⇒ $\cos\theta=\frac{1}{2}$
⇒ $\theta=60°$
Now, $\sec \theta+\operatorname{cosec} \theta+\cot \theta=\sec60°+\operatorname{cosec}60°+\cot60°$
$=2+\frac{2}{\sqrt3}+\frac1{\sqrt3}$
$=\frac{2\sqrt3+2+1}{\sqrt3}$
$=\frac{2\sqrt3+3}{\sqrt3}$
$=2+\sqrt3$
Hence, the correct answer is $2+\sqrt3$.
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