A) \[\sin {{18}^{o}}\]
B) \[2\sin {{18}^{o}}\]
C) \[2\cos {{18}^{o}}\]
D) \[2\cos {{36}^{o}}\]
Correct Answer: B
Solution :
Let\[\sin A=x\] ? (i) Then,\[\cos A=\tan B\] \[\Rightarrow \] \[\sqrt{1-{{\sin }^{2}}A}=\tan B\] \[\Rightarrow \] \[\sqrt{1-{{x}^{2}}}=\tan B\] Now, \[\cos B=\tan C\] \[\Rightarrow \] \[\frac{1}{\sqrt{1+{{\tan }^{2}}B}}=\tan C\] \[\Rightarrow \] \[\frac{1}{\sqrt{2-{{x}^{2}}}}=\tan C\] and \[\cos C=\frac{1}{\sqrt{1+{{\tan }^{2}}C}}=\sqrt{\frac{2-{{x}^{2}}}{3-{{x}^{2}}}}\] ...(ii) Also, \[\cos C=\tan A\] \[\Rightarrow \] \[\sqrt{\frac{2-{{x}^{2}}}{3-{{x}^{2}}}}=\frac{x}{\sqrt{1-{{x}^{2}}}}\] [From Eqs. (i) and (ii)] \[\Rightarrow \] \[{{x}^{2}}=\frac{{{(1\pm \sqrt{5})}^{2}}}{4}\] \[\Rightarrow \] \[x=\frac{\sqrt{5}-1}{2}=2\sin {{18}^{o}}\]
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