A) \[sin60{}^\circ \]
B) \[cos60{}^\circ \]
C) \[tan60{}^\circ \]
D) \[sin30{}^\circ \]
Correct Answer: A
Solution :
We have, \[\frac{2\tan \,30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }\] \[=\frac{2\times \left( \frac{1}{\sqrt{3}} \right)}{1+{{\left( \frac{1}{\sqrt{3}} \right)}^{2}}}\] \[\left[ \because \,\,\tan 30{}^\circ =\frac{1}{\sqrt{3}} \right]\] \[=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}\] \[=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{3}{\sqrt{3}}\times \frac{1}{2}\times \frac{\sqrt{3}}{\sqrt{3}}\] \[=\frac{3\times \sqrt{3}}{3\times 2}=\frac{\sqrt{3}}{2}=\sin \,60{}^\circ \]
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