The value of $$\frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} = ?$$
A. cos60°
B. tan60°
C. sin60°
D. sin30°
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} \cr & = \frac{{2 \times \sqrt 3 }}{{1 + 3}} \cr & = \frac{{\sqrt 3 }}{2} \cr & = \sin {60^ \circ } \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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