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 :
\[\tan A=1,\sin A=\frac{\tan A}{\sqrt{1+{{\tan }^{2}}A}}=\frac{1}{\sqrt{2}}\] \[\cos A=\frac{1}{\sqrt{1+{{\tan }^{2}}A}}=\frac{1}{\sqrt{2}}\] \[\tan B=\sqrt{3},\sin B=\frac{\sqrt{3}}{\sqrt{1+3}}=\frac{\sqrt{3}}{2}\] \[\cos B=\frac{1}{\sqrt{1+3}}=\frac{1}{2}\] \[\therefore \] \[\cos \,A\,.\,cos\,B\,-\,sin\,A\,.\,sin\,B\] \[=\frac{1}{\sqrt{2}}.\frac{1}{2}-\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2\sqrt{2}}\]
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