Solve the trigonometric equation $\tan \left( 5\theta \right)=\cot \left( 2\theta \right)$.

Hint: We will use the trigonometric formula given by $\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)$ to convert cotangent and tangent. We will also apply the formula $\tan \left( x \right)=\tan \left( y \right)$ results into $x=n\pi +y$ so that we can find the value of the angle which is required here.

Complete step-by-step answer:
We will first consider the equation $\tan \left( 5\theta \right)=\cot \left( 2\theta \right)$ as (i). Now, as we know that $\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)$ as the value of tangent is opposite to cotangent when it is in the first quadrant. Also, in the first quadrant the value of tan is positive. . Numerically we can write it as $\cot \left( 2\theta \right)=\tan \left( \dfrac{\pi }{2}-2\theta \right)$. Now we will substitute this value in equation (i). So we get $\tan \left( 5\theta \right)=\tan \left( \dfrac{\pi }{2}-2\theta \right)$.
Now we will proceed after it by the use of the formula given by $\tan \left( x \right)=\tan \left( y \right)$ results into $x=n\pi +y$. Now, we will apply this formula into the equation $\tan \left( 5\theta \right)=\tan \left( \dfrac{\pi }{2}-2\theta \right)$. Thus, we get
 $\begin{align}
  & \tan \left( 5\theta \right)=\tan \left( \dfrac{\pi }{2}-2\theta \right) \\
 & \Rightarrow 5\theta =n\pi +\dfrac{\pi }{2}-2\theta \\
\end{align}$
Now we will solve the above equation and we will have
$\begin{align}
  & 5\theta =n\pi +\dfrac{\pi }{2}-2\theta \\
 & \Rightarrow 5\theta +2\theta =\dfrac{\pi }{2}+n\pi \\
 & \Rightarrow 7\theta =\dfrac{\pi }{2}+n\pi \\
 & \Rightarrow \theta =\dfrac{\pi }{14}+\dfrac{n\pi }{7} \\
\end{align}$
Therefore the required value of the angle $\theta $ is given by $\theta =\dfrac{\pi }{14}+\dfrac{n\pi }{7}$.
Hence, the value of the trigonometric equation $\tan \left( 5\theta \right)=\cot \left( 2\theta \right)$ is given by $\theta =\dfrac{\pi }{14}+\dfrac{n\pi }{7}$.

Note: Alternate method of solving the trigonometric expression $\tan \left( 5\theta \right)=\cot \left( 2\theta \right)$ is given below.
We can write $\cot \left( 2\theta \right)$ as $\tan \left( \dfrac{\pi }{2}-2\theta \right)$. Therefore, we can use this value in the equation $\tan \left( 5\theta \right)=\cot \left( 2\theta \right)$. Therefore we have $\tan \left( 5\theta \right)=\tan \left( \dfrac{\pi }{2}-2\theta \right)$. Now we can take the term $\tan \left( \dfrac{\pi }{2}-2\theta \right)$ to the left side of the equation. Thus we have $\tan \left( 5\theta \right)-\tan \left( \dfrac{\pi }{2}-2\theta \right)=0$. And now we will write $\tan \left( 5\theta \right)$ as $\dfrac{\sin 5\theta }{\cos 5\theta }$. We will do the same with the term $\tan \left( \dfrac{\pi }{2}-2\theta \right)$ and write it as $\dfrac{\sin \left( \dfrac{\pi }{2}-2\theta \right)}{\cos \left( \dfrac{\pi }{2}-2\theta \right)}$. Now we will substitute the values into the trigonometric equation $\tan \left( 5\theta \right)-\tan \left( \dfrac{\pi }{2}-2\theta \right)=0$. Therefore we get a new equation now which is given by $\dfrac{\sin \left( 5\theta \right)}{\cos \left( 5\theta \right)}-\dfrac{\sin \left( \dfrac{\pi }{2}-2\theta \right)}{\cos \left( \dfrac{\pi }{2}-2\theta \right)}=0$. After taking lcm we will have, $\begin{align}
  & \dfrac{\sin \left( 5\theta \right)\cos \left( \dfrac{\pi }{2}-2\theta \right)-\sin \left( \dfrac{\pi }{2}-2\theta \right)\cos \left( 5\theta \right)}{\cos \left( 5\theta \right)\cos \left( \dfrac{\pi }{2}-2\theta \right)}=0 \\
 & \Rightarrow \sin \left( 5\theta \right)\cos \left( \dfrac{\pi }{2}-2\theta \right)-\sin \left( \dfrac{\pi }{2}-2\theta \right)\cos \left( 5\theta \right)=0 \\
\end{align}$
Now we will use the formula $\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$. Thus, now we have that $\begin{align}
  & \sin \left( 5\theta \right)\cos \left( \dfrac{\pi }{2}-2\theta \right)-\sin \left( \dfrac{\pi }{2}-2\theta \right)\cos \left( 5\theta \right)=0 \\
 & \Rightarrow \sin \left( 5\theta -\left( \dfrac{\pi }{2}-2\theta \right) \right)=0 \\
 & \Rightarrow \sin \left( 7\theta -\dfrac{\pi }{2} \right)=0 \\
\end{align}$
Now as we know that when $\sin x=0$ then $x=n\pi $. Therefore, we get $\sin \left( 7\theta -\dfrac{\pi }{2} \right)=0$ resulting into
$\begin{align}
  & 7\theta =n\pi +\dfrac{\pi }{2} \\
 & \Rightarrow \theta =\dfrac{n\pi }{7}+\dfrac{\pi }{14} \\
\end{align}$
Hence, the value of the trigonometric equation $\tan \left( 5\theta \right)=\cot \left( 2\theta \right)$ is given by $\theta =\dfrac{\pi }{14}+\dfrac{n\pi }{7}$.