A) \[\frac{1+\sqrt{3}}{2\sqrt{2}}\]
B) \[\frac{1-\sqrt{3}}{2\sqrt{2}}\]
C) \[\frac{2\sqrt{2}}{3}\]
D) 1
Correct Answer: B
Solution :
\[\because \]\[\tan A=1=\tan \frac{\pi }{4}\Rightarrow A=\frac{\pi }{4}\] and \[\tan B\sqrt{3}=\tan \frac{\pi }{3}\Rightarrow B=\frac{\pi }{3}\] \[\therefore \]\[\cos A.\cos \,\,B-sin\,\,A.sin\,\,B\] \[=cos\frac{\pi }{4}.\cos \frac{\pi }{3}-\sin \frac{\pi }{4}\sin \frac{\pi }{3}\] \[=\frac{1}{\sqrt{2}}\cdot \frac{1}{2}-\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2\sqrt{2}}\]
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