Question

Choose the correct option:
$\frac{2tan30^0}{1+ta{n}^2{30}^0}=$

[1 Mark]

(a) $sin{60}^{\circ }$
(b) $cos{60}^{\circ }$
(c) $tan{60}^{\circ }$
(d) $sin{30}^{\circ }$

Answer 1

The correct option is (a). [1 mark]
$\frac{2tan{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}$$=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+(\frac{1}{\sqrt{3}}{)}^2}$
$=\frac{\frac{2}{\sqrt{3}}}{(1+\frac{1}{3})}$$=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$=\frac{6}{4\sqrt{3}}=\frac{\sqrt{3}}{2}=sin{60}^{\circ }$