Question

Find the general solution of the expression $\tan p\theta =\cot q\theta $

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:


Given equation in the problem is

$\operatorname{tanp}\theta =cotq\theta .............\left( i \right)$

As we know tan and cot functions in terms of sin and cos can be given as

$\tan x=\dfrac{\sin x}{\cos x}..............\left( ii \right)$

And $\cot x=\dfrac{\cos x}{\sin x}................\left( iii \right)$

So, we can write the equation (i) in terms of sin and cos functions with the help of equations (ii) and (iii) as

$\dfrac{\operatorname{sinp}\theta }{\operatorname{cosp}\theta }=\dfrac{cosp\theta }{sinq\theta }$

On cross-multiplying the above equation, we get,

\[\operatorname{sinp}\theta \operatorname{sinq}\theta =\operatorname{cosp}\theta \operatorname{cosq}\theta \]

On transferring \[\operatorname{cosp}\theta \operatorname{cosq}\theta \] to the other side of the equation, we get

\[\begin{align}

  & \sin p\theta \sin q\theta -\cos p\theta \cos q\theta =0 \\

 & cosp\theta cosq\theta -\operatorname{sinp}\theta \operatorname{sinq}\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=p\theta $ and $B=q\theta $

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

$\begin{align}

  & \cos \left( p\theta +q\theta \right)=0 \\

 & \cos \left( \left( p+q \right)\theta \right)=0...............\left( vi \right) \\

\end{align}$

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

$x=\left( 2n+1 \right)\dfrac{\pi }{2}....................\left( vii \right)$

Where, $N\in Z$

So, the general solution of the equation (vi) with the help of above equation can be given as
$\left( p+q \right)\theta =\left( 2N+1 \right)\dfrac{\pi }{2}$

On dividing the whole equation (p + q), we get

$\begin{align}

  & \dfrac{\left( p+q \right)\theta }{\left( p+q \right)}=\dfrac{\left( 2N+1 \right)\pi }{2\left( p+q \right)} \\

 & \theta =\dfrac{\left( 2N+1 \right)\pi }{2\left( p+q \right)} \\

\end{align}$

Hence, the general solution of the given equation in the problem i.e. $\operatorname{tanp}\theta =\tan q\theta $ is given as

$\theta =\dfrac{\left( 2N+1 \right)\pi }{2\left( p+q \right)},where\text{ }n\in Z$

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

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

$\begin{align}

  & \tan p\theta =\cot q\theta \\

 & \tan p\theta =\tan \left( \dfrac{\pi }{2}-q\theta \right) \\

\end{align}$

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

$x=N\pi +y,$ where $N\in Z$

So, we get

$\begin{align}

  & p\theta =N\pi +\dfrac{\pi }{2}-q\theta \\

 & \left( p+q \right)\theta =\left( 2N+1 \right)\dfrac{\pi }{2} \\

 & \theta =\dfrac{\left( 2N+1 \right)\pi }{2\left( p+q \right)} \\

\end{align}$

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 try 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.

Be clear with the general solutions of $\sin x=0,\cos x=0$ and $\tan x=0$ . Don’t confuse with the general solutions of them. These can be given as

For $\sin x=0$

$x=n\pi ,n\in Z$

For $\cos x=0$

$x=\left( 2n+1 \right)\dfrac{\pi }{2},n\in Z$

For $\tan x=0$

$x=n\pi ,n\in Z$