Question

Prove that: $\tan 13\theta -\tan 9\theta -\tan 4\theta =\tan 13\theta \tan 9\theta \tan 4\theta $.

Answer 1

Hint: We will be using the concept of trigonometry to solve the problem. We will be using the fact that $\tan \left( 0{}^\circ \right)=0$, then we will write $0=13\theta -9\theta -4\theta $ and we apply the formula that \[\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\].

Complete step-by-step answer:

Now, we have to prove that,

$\tan 13\theta -\tan 9\theta -\tan 4\theta =\tan 13\theta \tan 9\theta \tan 4\theta $

Now, we know the fact that,

$\tan \theta {}^\circ =0$

Now, we can see that in the equation to prove we have tangents of $13\theta ,9\theta \ and\ 4\theta $. So, we will write zero in terms of $13\theta ,9\theta ,4\theta \ as\ \tan \left( 13\theta -9\theta -4\theta \right)=0$.

Now, we know the trigonometric formula that,

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

So, using this we have,

$\dfrac{\tan \left( 13\theta \right)+\tan \left( -9\theta \right)+\tan \left( -4\theta \right)-\tan \left( 13\theta \right)\tan \left( -9\theta \right)\tan \left( -4\theta \right)}{1-\tan \left( 13\theta \right)\tan \left( -9\theta \right)-\tan \left( -9\theta \right)\tan \left( -4\theta \right)-\tan \left( -4\theta \right)\tan \left( 13\theta \right)}=0$
Now, we know that,$\tan \left( -\theta \right)=-\tan \theta $ using this and on cross – multiplying we have,

$\tan \left( 13\theta \right)-\tan \left( 9\theta \right)-\tan \left( 4\theta \right)-\tan \left( 13\theta \right)\tan \left( 9\theta \right)\tan \left( 4\theta \right)=0$

Now, we will rearrange the terms to get,

$\tan \left( 13\theta \right)-\tan \left( 9\theta \right)-\tan \left( 4\theta \right)=\tan \left( 13\theta \right)\tan \left( 9\theta \right)\tan \left( 4\theta \right)$

Since, we have LHS = RHS.

Hence Proved.

Note: To solve these type of questions it is important to note that we have used the fact that $\tan \left( 0{}^\circ \right)=0$ and then by seeing the terms in equation to be proved we have written tan (0) as $\tan \left( 13\theta -9\theta -4\theta \right)$.