Question

If $A=30{}^\circ $, verify that: \[\sin 2A=\dfrac{2\tan A}{1+{{\tan }^{2}}A}\].

Answer 1

Hint: We will be using the concept of trigonometric functions to solve the problem. We will put the value of $A=30{}^\circ $ in LHS and RHS and further simplify each side to find the value of each side. Then we will show that since the value of both sides are equal for A therefore verified.

Complete step-by-step answer:
Now, we have been given that$A=30{}^\circ $.

We have to verify that \[\sin 2A=\dfrac{2\tan A}{1+{{\tan }^{2}}A}\].

Now, we will first take LHS which is given as $\sin 2A$ and then substitute the value of A in LHS to find its value. So, we have,

$\begin{align}

  & \sin 2A=\sin 2\times 30{}^\circ \\

 & =\sin 60{}^\circ \\

\end{align}$

Now, we know that the value of $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$. So, we have,

$\sin 2A=\dfrac{\sqrt{3}}{2}............\left( 1 \right)$

Now, we take RHS which have been given to us as \[\dfrac{2\tan A}{1+{{\tan }^{2}}A}\].

Now, we will substitute the value of $A=30{}^\circ $in the RHS. So, we have,

\[\dfrac{2\tan A}{1+{{\tan }^{2}}A}=\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }\]

Now, we know that the value of $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$. So, we have,

\[\begin{align}

  & \dfrac{2\tan A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}} \\

 & =\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}} \\

 & =\dfrac{2\times 3}{4\times \sqrt{3}} \\

 & \dfrac{2\tan A}{1+{{\tan }^{2}}A}=\dfrac{\sqrt{3}}{2}............\left( 2 \right) \\

\end{align}\]

Now, from (1) and (2) we have that,

\[\sin 2A=\dfrac{2\tan A}{1+{{\tan }^{2}}A}\]

Therefore, we have verified.


Note: To solve these type of question it is important to note that we have used a fact like,

$\begin{align}

  & \sin 60{}^\circ =\dfrac{\sqrt{3}}{2} \\

 & \tan 30{}^\circ =\dfrac{1}{\sqrt{3}} \\

\end{align}$

Also, it is important to notice that to verify the expression we have substituted the value of A in the expression.