Question : Using $\cos (A+B)=\cos A \cos B-\sin A \sin B$, find the value of $\cos 75^\circ$.
Option 1: $\frac{\sqrt{5}-1}{4}$
Option 2: $\frac{\sqrt{5}+1}{4}$
Option 3: $\frac{\sqrt{6}-\sqrt{2}}{4}$
Option 4: $\frac{\sqrt{6}+\sqrt{2}}{4}$
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Correct Answer: $\frac{\sqrt{6}-\sqrt{2}}{4}$
Solution : Given: $\cos (A+B)=\cos A \cos B-\sin A \sin B$ Let $A=45^\circ, B=30^\circ$ So, $\cos (45^\circ+30^\circ) = \cos 45^\circ \cos30^\circ - \sin45^\circ \sin 30^\circ$ ⇒ $\cos 75^\circ= \frac{1}{\sqrt{2}}×\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}×\frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ So, the value of $\cos 75^\circ$ is $\frac{\sqrt{6}-\sqrt{2}}{4}$. Hence, the correct answer is $\frac{\sqrt{6}-\sqrt{2}}{4}$.
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