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The value of $$\frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} = ?$$
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& \frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} \cr
& = \frac{{2 \times \sqrt 3 }}{{1 + 3}} \cr
& = \frac{{\sqrt 3 }}{2} \cr
& = \sin {60^ \circ } \cr} $$
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