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Question: If \(\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}\), then \(\tan 3\theta \) is equal to...

If tan⁡θ=1−cos⁡θsin⁡θ\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}, then tan⁡3θ\tan 3\theta is equal to

Explanation

Solution

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⁡θ\tan \theta . Now, to find the value of tan⁡3θ\tan 3\theta , we should know its formula. Substituting the value of tan⁡θ\tan \theta which we have found earlier in the formula of tan⁡3θ\tan 3\theta we give us our required answer of the same.

Complete step-by-step answer:
According to the question,
tan⁡θ=1−cos⁡θsin⁡θ\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}
Now, as we know,
cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta
⇒1−cos⁡2θ=2sin⁡2θ\Rightarrow 1 - \cos 2\theta = 2{\sin ^2}\theta…………………………..(1)\left( 1 \right)
Also,
sin⁡2θ=2sin⁡θ.cos⁡θ\sin 2\theta = 2\sin \theta .\cos \theta …………………………..(2)\left( 2 \right)
Now, using these half angle formulas i.e. (1)\left( 1 \right) and (2)\left( 2 \right) in the numerator and denominator of the given equation, we get,
tan⁡θ=2sin⁡2θ22sin⁡θ2.cos⁡θ2\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 2sin⁡θ22\sin \dfrac{\theta }{2} from numerator and denominator,
⇒tan⁡θ=sin⁡θ2cos⁡θ2\Rightarrow \tan \theta = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}
⇒tan⁡θ=tan⁡θ2\Rightarrow \tan \theta = \tan \dfrac{\theta }{2}…………………………………(3)\left( 3 \right)
(Because, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }})
Now, we have to find the value of tan⁡3θ\tan 3\theta .
As we know,
tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}
Now, substituting tan⁡θ\tan \theta by tan⁡θ2\tan \dfrac{\theta }{2} because of equation (3)\left( 3 \right) , we get,
tan⁡3θ=3tan⁡θ2−tan⁡3θ21−3tan⁡2θ2\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⁡3θ\tan 3\theta .
Therefore, we can say that,
If tan⁡θ=1−cos⁡θsin⁡θ\tan \theta = \dfrac{{1 - \cos \theta }}{{\sin \theta }}, then tan⁡3θ\tan 3\theta is equal to 3tan⁡θ2−tan⁡3θ21−3tan⁡2θ2\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⁡3θ\tan 3\theta . Otherwise, if we would try to prove the formula of tan⁡3θ\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.