A) \[cos{{60}^{{}^\circ }}\]
B) \[sin{{60}^{{}^\circ }}\]
C) \[tan{{60}^{{}^\circ }}\]
D) \[sin{{30}^{{}^\circ }}\]
Correct Answer: C
Solution :
(c) \[\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }\]\[=\frac{2\frac{1}{\sqrt{3}}}{1-{{\left( \frac{1}{\sqrt{3}} \right)}^{2}}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\sqrt{3}=\tan {{60}^{{}^\circ }}\]
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