Question

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

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

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

$\begin{align}

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

 & =\tan 60{}^\circ \\

\end{align}$

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

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

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

Now, we will substitute the value of A 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{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}} \\

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

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

\end{align}\]

Now, from (1) and (2) we have that if $A=30{}^\circ $ then,

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

Hence, verified.


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

$\begin{align}

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

 & \tan 60{}^\circ =\sqrt{3} \\

\end{align}$

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