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Question: The general solution of the equation \(\tan \theta =\tan \alpha \) is?...

The general solution of the equation tan⁡θ=tan⁡α\tan \theta =\tan \alpha is?

Explanation

Solution

Use the conversion formula tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} and take all the terms to the L.H.S. Now, take the L.C.M of the denominators of the terms in the L.H.S and in the numerator use the formula sin⁡acos⁡b−cos⁡asin⁡b=sin⁡(a−b)\sin a\cos b-\cos a\sin b=\sin \left( a-b \right) to simplify. Use the relation: if sin⁡x=0\sin x=0 then x=nπx=n\pi where ‘n’ is any integer. Form the relation between θ\theta and α\alpha by leaving θ\theta in the L.H.S and taking α\alpha to the R.H.S to get the answer.

Complete step by step answer:
Here we have been provided with the trigonometric equation tan⁡θ=tan⁡α\tan \theta =\tan \alpha and we are asked to find its general solution. Let us use the conversion tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} to simplify the equation first.
∵tan⁡θ=tan⁡α\because \tan \theta =\tan \alpha
⇒sin⁡θcos⁡θ=sin⁡αcos⁡α\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{\sin \alpha }{\cos \alpha }
Taking all the terms to the L.H.S we get,
⇒sin⁡θcos⁡θ−sin⁡αcos⁡α=0\Rightarrow \dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \alpha }{\cos \alpha }=0
Simplifying the terms by taking the L.C.M of the denominators of both the terms we get,

& \Rightarrow \dfrac{\sin \theta \cos \alpha -\sin \alpha \cos \theta }{\cos \theta \cos \alpha }=0 \\\ & \Rightarrow \sin \theta \cos \alpha -\sin \alpha \cos \theta =0 \\\ \end{aligned}$$ Using the trigonometric identity given as $\sin a\cos b-\cos a\sin b=\sin \left( a-b \right)$ we get, $$\Rightarrow \sin \left( \theta -\alpha \right)=0$$ Now, we know that the value of the sine function is 0 for any integral values of $\pi $, so we have the mathematical condition if $\sin x=0$ the $x=n\pi $ where n is any integer. Therefore we get the solution of the given equation as: - $$\begin{aligned} & \Rightarrow \left( \theta -\alpha \right)=n\pi \\\ & \therefore \theta =n\pi +\alpha ,n\in Z \\\ \end{aligned}$$ Hence, the above relation represents the general solution of the given equation. **Note:** You may note an important formula derived from the above solution which is given for the tangent function as if $\tan x=\tan y$ then we have $x=n\pi +y$. From here you can explain the fact that the tangent function has a period of $\pi $. You must remember the formulas of the general solutions of all the trigonometric equations like $\sin x=\sin y$ and $\cos x=\cos y$ because they will not be derived everywhere but directly used.