Question : If $(\cos \theta+\sin \theta):(\cos \theta-\sin \theta)=(\sqrt{3}+1):(\sqrt{3}-1), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\sec \theta$?
Option 1: 2
Option 2: $\sqrt{2}$
Option 3: 1
Option 4: $\frac{2 \sqrt{3}}{3}$
Correct Answer: $\frac{2 \sqrt{3}}{3}$
Solution :
$(\cos \theta+\sin \theta):(\cos \theta-\sin \theta)=(\sqrt{3}+1):(\sqrt{3}-1)$
$\Rightarrow \frac{\cos\ \theta\ +\ \sin\ \theta}{\cos\ \theta\ -\ \sin\ \theta}\ =\ \frac{\sqrt{3}+\ 1}{\sqrt{3}-\ 1}$
By componendo and dividendo
$\Rightarrow \frac{\cos\ \theta}{\sin\ \theta}\ =\ \frac{\sqrt{3}}{1}$
$\Rightarrow \cot\ \theta = \sqrt{3}=\cot30^\circ$
$\Rightarrow \theta = 30^\circ$
$\therefore \sec\ 30^\circ=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$
Hence, the correct answer is $\frac{2 \sqrt{3}}{3}$.
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