Answer
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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.
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.
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