A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[\frac{\sqrt{2}}{4}\]
Correct Answer: D
Solution :
| \[\therefore \] \[\cot A=\frac{1}{\sqrt{2}-1}\] |
 |
| \[\Rightarrow \]\[\tan A=\frac{\sqrt{2}-1}{1}\] |
| \[\therefore \] \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] |
| \[AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\sqrt{{{1}^{2}}+{{(\sqrt{2}-1)}^{2}}}\] |
| \[=\sqrt{1+2+1-2\sqrt{2}}=\sqrt{4-2\sqrt{2}}\] |
| \[\therefore \] \[\sin A=\frac{BC}{AC}=\frac{\sqrt{2}-1}{\sqrt{4}-2\sqrt{2}}\] |
| and \[\cos A=\frac{AB}{AC}=\frac{1}{\sqrt{4-2\sqrt{2}}}\] |
| \[\therefore \] \[\sin A\cos \,A=\frac{\sqrt{2}-1}{\sqrt{4}-2\sqrt{2}}\cdot \frac{1}{\sqrt{4}-2\sqrt{2}}\] |
| \[=\frac{\sqrt{2}-1}{\sqrt{4}-2\sqrt{2}}\] |
| \[=\frac{\sqrt{2}-1}{2\sqrt{2}\,(\sqrt{2}-1)}\] |
| \[=\frac{1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}\] |
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