Question

Prove that: $\dfrac{{{\tan }^{2}}2\theta -{{\tan }^{2}}\theta }{1-{{\tan }^{2}}2\theta {{\tan }^{2}}\theta }=\tan 3\theta -\tan \theta $.

Answer 1

Hint: We will be using the concept of trigonometry to solve the problem. We will be using the fact that,

$\begin{align}

  & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\

 & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\

\end{align}$,

Also, we will first apply algebraic identity that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, then we will use the formula for $\tan \left( A+B \right)\ and\ \tan \left( A-B \right)$.

Complete step-by-step answer:

Now, we have to prove that,

$\dfrac{{{\tan }^{2}}2\theta -{{\tan }^{2}}\theta }{1-{{\tan }^{2}}2\theta {{\tan }^{2}}\theta }=\tan 3\theta -\tan \theta $

Now, we will take LHS and prove it to be equal to RHS.

In LHS we have,

$\dfrac{{{\tan }^{2}}2\theta -{{\tan }^{2}}\theta }{1-{{\tan }^{2}}2\theta {{\tan }^{2}}\theta }$

Now, we will apply the algebraic identity that,

 ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$

So, we have,

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

Now, we know that,

$\begin{align}

  & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\

 & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\

\end{align}$

So, using this we have,

$\begin{align}

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

 & =\tan \left( 2\theta +\theta \right)\tan \left( 2\theta -\theta \right) \\

 & =\tan \left( 3\theta \right)\tan \left( \theta \right) \\

\end{align}$

Now, we have LHS = RHS. Hence proved.

Note: To solve these type of questions it is important to note that we have first applied an algebraic identity that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, then we have applied the formula that,

$\begin{align}

  & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\

 & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\

\end{align}$