Question

Prove that: $\tan 8\theta -\tan 6\theta -\tan 2\theta =\tan 8\theta \tan 6\theta \tan 2\theta $.

Answer 1

Hint: We will be using the concept of trigonometric ratio to solve the problem. We will using the trigonometric identity that \[\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\]. We will be first using the fact that $\tan \left( 0{}^\circ \right)=0$, then we will apply the trigonometric identity to solve the question.

Complete step-by-step answer:

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

In LHS we have,

$\tan 8\theta -\tan 6\theta -\tan 2\theta $

Now, we can see that,

$8\theta -6\theta -2\theta =0$

So, we will take tan on both sides. Therefore, we have,

$\tan \left( 8\theta -6\theta -2\theta \right)=\tan \theta $

Or we can write it as,

$\tan \theta =\tan \left( 8\theta -6\theta -2\theta \right)$

Now, we know that,

$\begin{align}

  & \tan \theta =0 \\

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

\end{align}$

So, we have,

$\theta =\dfrac{\tan \left( 8\theta \right)+\tan \left( -6\theta \right)+\tan \left( -2\theta \right)-\tan \left( 8\theta \right)\tan \left( -6\theta \right)\tan \left( -2\theta \right)}{1-\tan \left( 8\theta \right)\tan \left( -6\theta \right)-\tan \left( -6\theta \right)\tan \left( -2\theta \right)-\tan \left( -2\theta \right)\tan \left( 8\theta \right)}$

Now, we know that,

$\tan \left( -\theta \right)=-\tan \theta $

So, we have,

$\theta =\tan \left( 8\theta \right)-\tan \left( 6\theta \right)-\tan \left( 2\theta \right)-\tan 8\theta \tan 6\theta \tan 2\theta $

Now, we will rearrange the terms. So, we have,

$\tan 8\theta \tan 6\theta \tan 2\theta =\tan 8\theta -\tan 6\theta -\tan 2\theta $

Since, we have LHS = RHS.

Hence Proved.

Note: To solve these type of questions it is important to note that we have taken $8\theta -6\theta -2\theta =0$ and from there we have arrival at the equation that $\tan 8\theta -\tan 6\theta -\tan 2\theta =\tan 8\theta \tan 6\theta \tan 2\theta $. We have taken \[8\theta -6\theta -2\theta \] specifically as the sum of these terms is zero and also these terms are in the left hand side and right hand side of the equation to be proved.