Question

Choose the correct option $ \dfrac{2\tan {{30}^{\circ }}}{1+{{\tan }^{2}}{{30}^{\circ }}}= $ \[\]
A. $ \sin {{60}^{\circ }} $ \[\]
B. $ \cos {{60}^{\circ }} $ \[\]
C. $ \tan {{60}^{\circ }} $ \[\]
D. $ \sin {{30}^{\circ }} $ \[\]

Answer 1

Hint:We use the fact that tangent of any angle $ \theta $ is ratio of sine of the angle $ \theta $ to cosine of the angle $ \theta $ which means $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ . We find the value of $ \tan {{30}^{\circ }} $ using this ratio and put in the given expression to obtain the value. We compare the trigonometric values given in the options with the obtained value to find the correct option. \[\].

Complete step by step answer:
We know that in a right-angled triangle the side opposite to the right-angled triangle is called hypotenuse denoted as $ h $ , the vertical side is called perpendicular denoted as $ p $ and the horizontal side is called the base denoted as $ b $ .\[\]
We know from the trigonometric ratios in a right-angled triangle the sine of any angle is given by the ratio of the side opposite to the angle to the hypotenuse. In the figure, the sine of the angle $ \theta $ is given by
$\Rightarrow$ \[\sin \theta =\dfrac{p}{h}...(1)\]
Similarly the cosine of an angle is the ratio of the side adjacent to the angle (excluding hypotenuse) to the hypotenuse. So we have cosine of angle $ \theta $
$\Rightarrow$ \[\cos \theta =\dfrac{b}{h}...(2)\]
The tangent of the angle is the ratio of opposite side to the adjacent side(excluding hypotenuse) . So we have tangent of the angle of angle $ \theta $
$\Rightarrow$ \[\tan \theta =\dfrac{p}{b}\]
Let us divide numerator and denominator by $ h $ and use the values obtained in (1) and (2). So we have,
$\Rightarrow$ \[\tan \theta =\dfrac{\dfrac{p}{h}}{\dfrac{b}{h}}=\dfrac{\sin \theta }{\cos \theta }\]
We also know the sine and cosine values of $ {{30}^{\circ }} $ . We have
$\Rightarrow$ \[\sin {{30}^{\circ }}=\dfrac{1}{2},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
So the tangent of $ {{30}^{\circ }} $ is ,
 $ \Rightarrow $ \[\tan {{30}^{\circ }}=\dfrac{\sin {{30}^{\circ }}}{\cos {{30}^{\circ }}}=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{1}{\sqrt{3}}\]
The given expression is
 $ \Rightarrow $ \[\dfrac{2\tan {{30}^{\circ }}}{1+{{\tan }^{2}}{{30}^{\circ }}}\]
We substitute the obtained value of $ \tan {{30}^{\circ }} $ and simplify . So we have,
\[\dfrac{2\tan {{30}^{\circ }}}{1+{{\tan }^{2}}{{30}^{\circ }}}=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}=\dfrac{\sqrt{3}}{2}\]
We know that $ \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2},\cos {{60}^{\circ }}=\dfrac{1}{2},\tan {{60}^{\circ }}=\sqrt{3},\sin {{30}^{\circ }}=\dfrac{1}{2} $ . So
 $ \Rightarrow $ \[\dfrac{2\tan {{30}^{\circ }}}{1+{{\tan }^{2}}{{30}^{\circ }}}=\sin {{60}^{\circ }}\]
So the correct option is A
Note:
We can directly find the answer using the double angle formula of sine in terms of the tangent which is $ \sin 2A=\dfrac{2\tan A}{1+{{\tan }^{2}}A} $ . While using it we must be careful of the confusion from tangent double angle formula which is $ \tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A} $ .