Question

If $\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}$, then $\tan 3\theta $ is equal to

Answer 1

Hint: To solve this question, we would use half angle formulas in the numerator as well as the denominator of RHS. Then, cancelling out the same terms would give us the value of $\tan \theta $ . Now, to find the value of $\tan 3\theta $ , we should know its formula. Substituting the value of $\tan \theta $ which we have found earlier in the formula of $\tan 3\theta $ we give us our required answer of the same.

Complete step-by-step answer:
According to the question,
$\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}$
Now, as we know,
$\cos 2\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow 1 - \cos 2\theta = 2{\sin ^2}\theta $…………………………..$\left( 1 \right)$
Also,
$\sin 2\theta = 2\sin \theta .\cos \theta $…………………………..$\left( 2 \right)$
Now, using these half angle formulas i.e. $\left( 1 \right)$ and $\left( 2 \right)$ in the numerator and denominator of the given equation, we get,
$\tan \theta = \dfrac{{2{{\sin }^2}\dfrac{\theta }{2}}}{{2\sin \dfrac{\theta }{2}.\cos \dfrac{\theta }{2}}}$
These formulas are called half angle formulas because when we apply them, they reduce the given angle to its half angle (as we saw in this question as well).
Now, we would cancel out $2\sin \dfrac{\theta }{2}$ from numerator and denominator,
$ \Rightarrow \tan \theta = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}$
$ \Rightarrow \tan \theta = \tan \dfrac{\theta }{2}$…………………………………$\left( 3 \right)$
(Because, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$)
Now, we have to find the value of $\tan 3\theta $.
As we know,
$\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$
Now, substituting $\tan \theta $ by $\tan \dfrac{\theta }{2}$ because of equation $\left( 3 \right)$ , we get,
$\tan 3\theta = \dfrac{{3\tan \dfrac{\theta }{2} - {{\tan }^3}\dfrac{\theta }{2}}}{{1 - 3{{\tan }^2}\dfrac{\theta }{2}}}$
Now, this cannot be solved further, and hence, this is the required value of $\tan 3\theta $.
Therefore, we can say that,
If $\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}$, then $\tan 3\theta $ is equal to $\dfrac{{3\tan \dfrac{\theta }{2} - {{\tan }^3}\dfrac{\theta }{2}}}{{1 - 3{{\tan }^2}\dfrac{\theta }{2}}}$.
Hence, this is our required answer.

Note: In order to answer these questions, we should know the half angle formulas and specially the formula of $\tan 3\theta $ . Otherwise, if we would try to prove the formula of $\tan 3\theta $ then unnecessary it would consume a lot of time. Hence, knowing these formulas and how to imply them would help us reach our required answer.