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