Question

Prove the following: $\tan 4\theta =\dfrac{4\tan \theta \left( 1-{{\tan }^{2}}\theta \right)}{1-6{{\tan }^{2}}\theta +{{\tan }^{4}}\theta }$.

Answer 1

Hint: To prove the expression given in the question, we need to know that $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$, which is derived using the relation, $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x+\tan y}$ where $x=y=\theta $.

Complete step-by-step answer:

In this question, we are asked to prove that $\tan 4\theta =\dfrac{4\tan \theta \left( 1-{{\tan }^{2}}\theta \right)}{1-6{{\tan }^{2}}\theta +{{\tan }^{4}}\theta }$. In order to solve that we will first consider the left hand side (LHS) of the given expression, that is, $\tan 4\theta $.

We know that $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$, so we will use it to simplify $\tan 4\theta $. So, we will get, $\tan 4\theta =\tan 2\left( 2\theta \right)$
$\Rightarrow \tan 4\theta =\dfrac{2\tan \left( 2\theta \right)}{1-{{\tan }^{2}}\left( 2\theta \right)}$

Now, we will again substitute the value of $\tan 2\theta $, that is $\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ in the above expression. So we get,

$\tan 4\theta =\dfrac{2\left[ \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right]}{1-{{\left[ \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right]}^{2}}}$

On further simplification, we will get,

$\tan 4\theta =\dfrac{2\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)}{1-\dfrac{{{\left( 2\tan \theta \right)}^{2}}}{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}}$

Now, we will take the LCM of the denominator. So, we will get the expression as,

$\tan 4\theta =\dfrac{\dfrac{2\left( \tan \theta \right)}{\left( 1-{{\tan }^{2}}\theta \right)}}{\dfrac{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-{{\left( 2\tan \theta \right)}^{2}}}{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}}$

The above expression can also be written as follows:

$\tan 4\theta =\dfrac{\left( 4\tan \theta \right){{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}{\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]\left( 1-{{\tan }^{2}}\theta \right)}$

Now, we can see that $\left( 1-{{\tan }^{2}}\theta \right)$ is common in the numerator and in the denominator, so we can cancel them. Also, we know that ${{\left( a-b \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}-2ab \right)$, so we can expand the term $\left( 1-{{\tan }^{2}}\theta \right)$ and write it as $\left( 1+{{\tan }^{4}}\theta -2{{\tan }^{2}}\theta \right)$. So, we get,

$\tan 4\theta =\dfrac{\left( 4\tan \theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1+{{\tan }^{4}}\theta -2{{\tan }^{2}}\theta \right)-4{{\tan }^{2}}\theta }$

Now, we can simplify the above expression further, so we get,

$\begin{align}

  & \tan 4\theta =\dfrac{\left( 4\tan \theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{1+{{\tan }^{4}}\theta -2{{\tan }^{2}}\theta -4{{\tan }^{2}}\theta } \\

 & \Rightarrow \tan 4\theta =\dfrac{4\tan \theta \left( 1-{{\tan }^{2}}\theta \right)}{1+{{\tan }^{4}}\theta -6{{\tan }^{2}}\theta } \\

 & \Rightarrow \tan 4\theta =\dfrac{4\tan \theta \left( 1-{{\tan }^{2}}\theta \right)}{1-6{{\tan }^{2}}\theta +{{\tan }^{4}}\theta } \\

\end{align}$

Therefore, we get the LHS = RHS.

Hence, we have proved that $\tan 4\theta =\dfrac{4\tan \theta \left( 1-{{\tan }^{2}}\theta \right)}{1-6{{\tan }^{2}}\theta +{{\tan }^{4}}\theta }$.

Note: We can also prove this question using the relation $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x+\tan y}$, where $x=\theta $ and $y=3\theta $. And then, again we will apply the same formula for $3\theta =\theta +2\theta $. And then, will continue applying until we get each term of the numerator as well as the denominator in terms of $\tan \theta $. This solution will get complicated and lengthier, hence this method is not preferred and it is advised to solve this question using the relation, $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$.