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Question: Solve \(\tan 3\theta = \cot \theta \) ?...

Solve tan⁡3θ=cot⁡θ\tan 3\theta = \cot \theta ?

Explanation

Solution

We can solve this question by applying the trigonometric formula of tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} , where θ\theta are the angles . In the fourth quadrant the tangent and cotangent remains positive along with first quadrant where all the wanted sine, cosine, tangent remains positive. Also remember the trigonometry table so that we can quickly assign the values as per the angles , and will find the general solution of tan⁡θ=tan⁡α\tan \theta = \tan \alpha which will be further used in solving the given question .

Complete step by step answer:
We are given the question for solving tan⁡3θ=cot⁡θ\tan 3\theta = \cot \theta . We are going to solve it by applying the trigonometric values corresponding to the angles .
tan⁡3θ=cot⁡θ\tan 3\theta = \cot \theta
As we know that there is a property in trigonometry that tan⁡(90∘−θ)=cot⁡θ\tan ({90^ \circ } - \theta ) = \cot \theta . So , substituting the cot θ\theta value we get –
tan⁡3θ=tan⁡(π2−θ)\tan 3\theta = \tan (\dfrac{\pi }{2} - \theta )---------- eq1
Now , in order to get the general solution the steps are –
tan⁡θ=tan⁡α\tan \theta = \tan \alpha
By applying the trigonometric formula of tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
sin⁡θcos⁡θ=sin⁡αcos⁡α\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\sin \alpha }}{{\cos \alpha }}
sin⁡θcos⁡α=sin⁡αcos⁡θ\sin \theta \cos \alpha = \sin \alpha \cos \theta

As we know that there is formula in trigonometry that sin⁡θcos⁡α−sin⁡αcos⁡θ=sin⁡(θ−α)\sin \theta \cos \alpha - \sin \alpha \cos \theta = \sin (\theta - \alpha )
Substituting the same we get –
sin⁡(θ−α)=0\sin (\theta - \alpha ) = 0
With the help of trigonometric table we can say sin⁡180∘=0\sin {180^ \circ } = 0 –
sin⁡x=0 x=nπ,n∈Z  \sin x = 0 \\\ x = n\pi ,n \in \mathbb{Z} \\\
Similarly , for sin⁡(θ−α)=0\sin (\theta - \alpha ) = 0
θ−α=nπ,n∈Z θ=nπ+α,n∈Z  \theta - \alpha = n\pi ,n \in \mathbb{Z} \\\ \theta = n\pi + \alpha ,n \in \mathbb{Z} \\\

Now coming back to eq. 1 , we will solve further as we have the proof of general solution θ=nπ+α,n∈Z\theta = n\pi + \alpha ,n \in \mathbb{Z}
tan⁡3θ=tan⁡(π2−θ)\tan 3\theta = \tan (\dfrac{\pi }{2} - \theta )
Considering (π2−θ)(\dfrac{\pi }{2} - \theta ) as α\alpha , we get
3θ=nπ+(π2−θ) 3θ=nπ+π2−θ 4θ=nπ+π2 θ=nπ4+π8 θ=(2n+1)π8  3\theta = n\pi + (\dfrac{\pi }{2} - \theta ) \\\ 3\theta = n\pi + \dfrac{\pi }{2} - \theta \\\ 4\theta = n\pi + \dfrac{\pi }{2} \\\ \theta = \dfrac{{n\pi }}{4} + \dfrac{\pi }{8} \\\ \theta = (2n + 1)\dfrac{\pi }{8} \\\
Therefore, the required solution θ=(2n+1)π8\theta = (2n + 1)\dfrac{\pi }{8}.

Note: One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer .
Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus .
Remember the Sum and difference Formulae and Reciprocal identities as per the application and usage .
We should know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin \left( { - \theta } \right) = - \sin \theta .\cos \left( { - \theta } \right) = \cos \theta \,and\,\tan \left( { - \theta } \right) = - \tan \theta
As a matter of fact, sin⁡θ\sin \theta andtan⁡θ\tan \theta and their reciprocals, csc⁡θ\csc \theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta s aere even functions .