Question : If $a=45^{\circ}$ and $b=15^{\circ}$, what is the value of $\frac{\cos (a-b)-\cos (a+b)}{\cos (a-b)+\cos (a+b)} ?$
Option 1: $2-2 \sqrt{2}$
Option 2: $3-\sqrt{6}$
Option 3: $3-\sqrt{2}$
Option 4: $2-\sqrt{3}$
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Correct Answer: $2-\sqrt{3}$
Solution :
Given, $a=45°$ and $b=15°$
$a-b = 45° - 15° = 30°$
And, $a+b = 45° + 15° = 60°$
So, $\frac{\cos (a-b)-\cos (a+b)}{\cos (a-b)+\cos (a+b)}$
$= \frac{\cos(30°) - \cos(60°)}{\cos(30°) + \cos(60°)}$
$= \frac{\frac{\sqrt{3}}{2} - \frac{1}{2}}{\frac{\sqrt{3}}{2}+\frac{1}{2}}$
⇒ $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3}+1}\times \frac{\sqrt{3} - 1}{\sqrt{3}+1}$
$= \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 - 2\sqrt{3}}{3-1}$
⇒ $\frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$
Hence, the correct answer is $2-\sqrt{3}$.
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