Question

Solve the given trigonometric expression
$\tan 3\theta = \cot \theta$

Answer 1

Hint- In this question convert $\tan \theta$ in terms of $\sin \theta {\text{ and cos}}\theta$ and similarly $\cot \theta$ in terms of $\sin \theta {\text{ and cos}}\theta$ using the concept that tan is the ratio of sine to cosine and cot is the ratio of cosine to sine. Use the general formula for roots when $\cos \theta = 0$, this will help getting the right answer.

Complete step-by-step solution -

Given trigonometric equation
$\tan 3\theta = \cot \theta$
As we know tan is the ratio of sine to cosine and cot is the ratio of cosine to sine so use this property we have,
$\Rightarrow \dfrac{{\sin 3\theta }}{{\cos 3\theta }} = \dfrac{{\cot \theta }}{{\sin \theta }}$
Now simplify this we have,
$\Rightarrow \sin 3\theta \sin \theta = \cos 3\theta \cos \theta$
$\Rightarrow \cos 3\theta \cos \theta - \sin 3\theta \sin \theta = 0$
Now as we know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ so use this property we have,
$\Rightarrow \cos \left( {3\theta + \theta } \right) = 0$
$\Rightarrow \cos 4\theta = 0$
Now as we know that $0 = \cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$, where n = 0, 1, 2...
$\Rightarrow \cos 4\theta = 0 = \cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$
Now on comparing we have,
$\Rightarrow 4\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}$
Now divide by 4 throughout we have,
$\Rightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{8}$, $n \in 0,1,2........$
So this is the required solution of the given expression.

Note – The verification of $\cos 4\theta = 0 = \cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$ can be provided by the fact that n in this case varies form n = 0, 1, 2... that is set of whole numbers. So if n=0, then we will have $\cos \dfrac{\pi }{2}$which eventually will be zero. Now if we substitute 1 in place of n we get $\cos \dfrac{{3\pi }}{2}$which is again zero. Thus $\cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$ is the general value for any $\cos \theta = 0$.