Question

Find the general solution of the expression $\tan m\theta +\cot n\theta =0$

Answer 1

Hint: Use the relations: $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ and \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \] with the given expression to get the whole equation in terms of sine and cosine functions only. Apply $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ identity to get a simplified form of the given equation. General solution of $\cos x=0$ is $x=\left( 2n+1 \right)\dfrac{\pi }{2},$ where $n\in Z;$

Complete step-by-step answer:

$\dfrac{\sin m\theta }{\cos m\theta }+\dfrac{\cos n\theta }{\sin n\theta }=0$

Now, taking L.C.M of denominators of both fractions, we get

$\dfrac{\sin m\theta \sin n\theta +\cos n\theta \cos m\theta }{\cos m\theta \sin n\theta }=\dfrac{0}{1}$

On cross-multiplying the above equation, we get

$\begin{align}

  & \sin m\theta \sin n\theta +\cos n\theta \operatorname{cosm}\theta =0 \\

 & \Rightarrow \cos n\theta \operatorname{cosm}\theta +\operatorname{sinn}\theta \operatorname{sinm}\theta =0...............\left( iv \right) \\

\end{align}$

As we know the trigonometric identity of $\cos \left( A+B \right)$ is given as

$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B..............\left( v \right)$

Now, we can compare the equation (iv) with the right-hand side of the equation (v). So, we get

$A=n\theta $ and $B=m\theta $

So, we can re-write the equation (iv) as

$\begin{align}

  & \cos \left( n\theta -m\theta \right) \\

 & \Rightarrow \cos \left( n-m \right)\theta =0...............\left( vi \right) \\

\end{align}$

Now, we know the general solution of $\cos x=0$ can be given as

\[x=\left( 2N+1 \right)\dfrac{\pi }{2}....................\left( vii \right)\]

So, the general solution of the equation (vi) with the help of above equation can be given as

$\begin{align}

  & n\theta -m\theta =\dfrac{\left( 2N+1 \right)\pi }{2} \\

 & \theta =\dfrac{\left( 2N+1 \right)\pi }{2\left( n-m \right)} \\

\end{align}$

where, $N\in Z$

Hence, the general solution of the given equation in the problem is given as $\theta =\dfrac{\left( 2N+1 \right)\pi }{2\left( n-m \right)}$ , where $n\in Z$.

So, we can put ‘n’ as integers to get the required solutions.

Note: Another approach to get general solution for the given expression would be

$\begin{align}

  & \tan m\theta +\cot n\theta =0 \\

 & \tan m\theta =-\cot n\theta \\

 & \tan m\theta =\tan \left( \dfrac{\pi }{2}+n\theta \right) \\

\end{align}$

Now, use the general solution for $\tan x=\tan y$ , which is given as

$\begin{align}

  & x=N\pi +y,N\in Z \\

 & m\theta =n\pi +\dfrac{\pi }{2}+n\theta \\

 & m\theta -n\theta =\left( 2N+1 \right)\dfrac{\pi }{2} \\

 & \theta =\dfrac{\left( 2N+1 \right)\pi }{2\left( m-n \right)} \\

\end{align}$

where, $N\in Z$

One may go wrong if/she uses the solution of equation, $\cos x=0$ as $x=\left( 2n+1 \right)\dfrac{\pi }{2},$ as ‘n’ variable is already involved in the given question, so write the general solution of $\cos x=0$ in terms of any other variable to not confuse yourself. So, take care of it for future reference as well.