Question

Solve the equation \[\tan 3\theta +\tan \theta =2\tan 2\theta \].

Answer 1

Hint: As we know that the tangent of any angle is equal to the ratio of sine to cosine of that angle, so we will substitute the value of tangent in terms of sine and cosine. Also we will use the trigonometric identity which is given as follows:

Complete step-by-step answer:

\[\sin A\cos B\pm \cos A\sin B=\sin \left( A\pm B \right)\]

We have been given the equation \[\tan 3\theta +\tan \theta =2\tan 2\theta \].

\[\begin{align}

  & \tan 3\theta +\tan \theta =\tan 2\theta +\tan 2\theta \\

 & \tan 3\theta +\tan 2\theta =\tan 2\theta -\tan \theta \\

\end{align}\]

As we know that tangent of any angle is equal to the ratio of sine to cosine of that angle.

So by substituting the value of tangent in terms of sine and cosine, we get as follows:

\[\dfrac{\sin 3\theta }{\cos 3\theta }-\dfrac{\sin 2\theta }{\cos 2\theta }=\dfrac{\sin 2\theta }{\cos 2\theta }-\dfrac{\sin \theta }{\cos \theta }\]

Taking LCM of terms on both the sides of the equality, we get as follows:

\[\dfrac{\sin 3\theta \cos 2\theta -\sin 2\theta \cos 3\theta }{\cos 3\theta \cos 2\theta }=\dfrac{\sin 2\theta \cos \theta -\sin \theta \cos 2\theta }{\cos 2\theta \cos \theta }\]

As we know that \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\].

So by using the trigonometric identity mentioned above, we get as follows:

\[\begin{align}

  & \dfrac{\sin \left( 3\theta -2\theta \right)}{\cos 3\theta \cos 2\theta }=\dfrac{\sin \left( 2\theta -\theta \right)}{\cos 2\theta \cos \theta } \\

 & \dfrac{\sin \left( \theta \right)}{\cos 3\theta \cos 2\theta }-\dfrac{\sin \left( \theta \right)}{\cos 2\theta \cos \theta }=0 \\

\end{align}\]

Taking \[\sin \theta \] common, we get as follows:

\[\begin{align}

  & \sin \theta \left[ \dfrac{1}{\cos 3\theta \cos 2\theta }-\dfrac{1}{\cos 2\theta \cos \theta } \right]=0 \\

 & \dfrac{\sin \theta }{\cos 2\theta }\left[ \dfrac{1}{\cos 3\theta }-\dfrac{1}{\cos \theta } \right]=0 \\

 & \dfrac{\sin \theta }{\cos 2\theta }\left[ \dfrac{\cos \theta -\cos 3\theta }{\cos \theta \cos 3\theta } \right]=0 \\

\end{align}\]

As we know that \[\cos C-\cos D=-\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\].

So, by using the trigonometric identity mentioned above, we get as follows:

\[\begin{align}

  & \dfrac{\sin \theta }{\cos \theta \cos 2\theta \cos 3\theta }\left[ -2\sin \dfrac{\theta +3\theta }{2}.\sin \dfrac{\theta -3\theta }{2} \right]=0 \\

 & \dfrac{\sin \theta }{\cos \theta \cos 2\theta \cos 3\theta }\left[ -2\sin \dfrac{4\theta }{2}\sin \dfrac{-2\theta }{2} \right]=0 \\

 & \dfrac{\sin \theta }{\cos \theta \cos 2\theta \cos 3\theta }\left[ -2\sin 2\theta .\sin (-\theta ) \right]=0 \\

\end{align}\]

Sine we know that \[\sin (-x)=-\sin x\], we can rewrite the equation as,

\[\dfrac{\sin \theta }{\cos \theta \cos 2\theta \cos 3\theta }\left[ -2\sin 2\theta .-\sin (\theta ) \right]=0\]

By simplifying the above equation, we can equate:

\[\Rightarrow {{\sin }^{2}}\theta =0\] and \[\sin 2\theta =0\].

As we know that the general solution for \[{{\sin }^{2}}\theta ={{\sin }^{2}}\alpha \] is \[\theta =n\pi +\alpha \] and for \[\sin \theta =\sin \alpha \] is \[\theta =2n\pi +\alpha \].

\[\Rightarrow \theta =n\pi \] and \[2\theta =n\pi \Rightarrow \theta =\dfrac{n\pi }{2}\].

But \[\theta =\dfrac{n\pi }{3}\] is rejected when n is odd as it makes \[\infty =\infty \].

Therefore, \[\theta =n\pi \] is the solution of the given equation as it makes 0 = 0.


Note: Check the final solution by substituting these values in the given equation.

Also, be careful while substituting the trigonometric identity in the equation as there is a chance that you might put the wrong sign.