Question

Choose the correct option and justify your choice: $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=$
(A) $\sin 60{}^\circ $
(B) $\cos 60{}^\circ $
(C) $\tan 60{}^\circ $
(D) $\sin 30{}^\circ $

Answer 1

Hint: In this question we will just put the value of $\tan 30{}^\circ $ in the equation and we will try to simplify the equation. We will take L.C.M in the denominator and then we will take the denominator part in the multiplication and then we will cut the common terms and will find the final answer.

Complete step by step solution:
The given equation is $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$. We know that the value of $\tan 30{}^\circ $ is $\dfrac{1}{\sqrt{3}}$. Now we will put the value of $\tan 30{}^\circ $ in the equation $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$.
Now we can write $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$ as follows and we will simplify it.
$\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$
Now, find the square of $\dfrac{1}{\sqrt{3}}$
$=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}$
Now take the L.C.M in denominator. Therefore, the above equation we look like:
\[=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}\]
Now, just bring the denominator part into multiplication. Therefore, the equation will look like:
$=\dfrac{2}{\sqrt{3}}\times \dfrac{3}{4}$
Now, we can write $3$ as $\sqrt{3}\times \sqrt{3}$ in the above equation. Therefore, the equation will look like:
$=\dfrac{2}{\sqrt{3}}\times \dfrac{\sqrt{3}\times \sqrt{3}}{4}$
Now, just cancel out the common terms and then the above equation will look like:
$=\dfrac{\sqrt{3}}{2}$
Now, we know that the value of $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$. The value of $\cos 60{}^\circ $ is$\dfrac{1}{2}$. The value of $\tan 60{}^\circ $ is $\sqrt{3}$. The value of $\sin 30{}^\circ $ is $\dfrac{1}{2}$. Therefore, we can say that the answer is $\sin 60{}^\circ $.

Hence, option $(\text{A)}$ is the correct option.

Note: The important thing is this question is we should be able to recall the value of $\tan 30{}^\circ $ because if we don’t know its value that we will not be able to solve this type of question. So just be careful about it and memorize the required values. The manipulations which we have done in one of the steps in this question are also important.
The values trigonometric functions can be seen in below table:

Trigonometry Ratios Table
Angles (In Degree)	\[0{}^\circ \]	\[30{}^\circ \]	\[45{}^\circ \]	\[60{}^\circ \]	\[90{}^\circ \]	\[180{}^\circ \]	\[270{}^\circ \]	\[360{}^\circ \]
Angles(In Radians)	\[0{}^\circ \]	\[\dfrac{\pi }{6}\]	\[\dfrac{\pi }{4}\]	\[\dfrac{\pi }{3}\]	\[\dfrac{\pi }{2}\]	\[\pi \]	\[\dfrac{3\pi }{2}\]	\[2\pi \]
sin	\[0\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{\sqrt{3}}{2}\]	\[1\]	\[0\]	\[-1\]	\[0\]
cos	\[1\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{2}\]	\[0\]	\[-1\]	\[0\]	\[1\]
tan	\[0\]	\[\dfrac{1}{\sqrt{3}}\]	\[1\]	\[\sqrt{3}\]	\[\infty \]	\[0\]	\[\infty \]	\[0\]
cot	\[\infty \]	\[\sqrt{3}\]	\[1\]	\[\dfrac{1}{\sqrt{3}}\]	\[0\]	\[\infty \]	\[0\]	\[\infty \]
cosec	\[\infty \]	\[2\]	\[\sqrt{2}\]	\[\dfrac{2}{\sqrt{3}}\]	\[1\]	\[\infty \]	\[-1\]	\[\infty \]
sec	\[1\]	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	\[2\]	\[\infty \]	\[-1\]	\[\infty \]	\[1\].