A) 1
B) \[\sqrt{3}\]
C) \[\frac{\sqrt{3}}{2}\]
D) 2
Correct Answer: C
Solution :
\[\cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] \[\therefore \frac{1-{{\tan }^{2}}{{15}^{o}}}{1+{{\tan }^{2}}{{15}^{o}}}=\cos {{30}^{o}}=\frac{\sqrt{3}}{2}\] \[[\because \theta ={{15}^{o}}]\]
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