A) \[\frac{1}{\sqrt{3}}\]
B) \[\sqrt{3}\]
C) 1
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
Let sides of a square be a. |
Then, \[AC=a\sqrt{2}\] and \[AO=OC=\frac{a}{\sqrt{2}}\] |
Here, \[AM=\frac{a}{2}\] |
\[\therefore \] \[LM=\frac{a}{\sqrt{2}}-\frac{a}{2}\] and \[OM=\frac{a}{2}\] |
In \[\Delta OML,\] |
\[\tan \frac{\theta }{2}=\frac{\frac{a}{\sqrt{2}}-\frac{a}{2}}{\frac{a}{2}}=\frac{\frac{\sqrt{2}-1}{2}}{\frac{1}{2}}=\sqrt{2}-1\] |
\[\therefore \] \[\tan \theta =\frac{2\tan \frac{\theta }{2}}{1-{{\tan }^{2}}\frac{\theta }{2}}=\frac{2(\sqrt{2}-1)}{1-(2+1-2\sqrt{2})}\] |
\[=\frac{2(\sqrt{2}-1)}{1-3+2\sqrt{2}}=\frac{2\,\,(\sqrt{2}-1)}{2\sqrt{2}-2}\]\[\Rightarrow \]\[\tan \theta =1\] |
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