Question

The general solution of the equation $\tan \theta =\tan \alpha $ is?

Answer 1

Hint: Use the conversion formula $\tan x=\dfrac{\sin x}{\cos x}$ 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 $\sin a\cos b-\cos a\sin b=\sin \left( a-b \right)$ to simplify. Use the relation: if $\sin x=0$ then $x=n\pi $ where ‘n’ is any integer. Form the relation between $\theta $ and $\alpha $ by leaving $\theta $ in the L.H.S and taking $\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 \theta =\tan \alpha $ and we are asked to find its general solution. Let us use the conversion $\tan x=\dfrac{\sin x}{\cos x}$ to simplify the equation first.
$\because \tan \theta =\tan \alpha $
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{\sin \alpha }{\cos \alpha }$
Taking all the terms to the L.H.S we get,
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \alpha }{\cos \alpha }=0$
 Simplifying the terms by taking the L.C.M of the denominators of both the terms we get,
\[\begin{align}
  & \Rightarrow \dfrac{\sin \theta \cos \alpha -\sin \alpha \cos \theta }{\cos \theta \cos \alpha }=0 \\
 & \Rightarrow \sin \theta \cos \alpha -\sin \alpha \cos \theta =0 \\
\end{align}\]
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: -
\[\begin{align}
  & \Rightarrow \left( \theta -\alpha \right)=n\pi \\
 & \therefore \theta =n\pi +\alpha ,n\in Z \\
\end{align}\]
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.