If $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\cos \left( \dfrac{\pi }{4}\tan \theta \right)$ then $\theta =$.
A. \[n\pi +\dfrac{\pi }{4}\]
B. \[2n\pi \pm \dfrac{\pi }{4}\]
C. \[n\pi -\dfrac{\pi }{4}\]
D. \[2n\pi \pm \dfrac{\pi }{6}\]

Hint: To find the value of $\theta $ we will first write the given equation in terms of sin. Then comparing both side of the equation we will derive an equation which we will simplify. We will then write the derived equation in terms of tan and again simplify using formula of expansion. We will then apply the theorem which states that if $x,y$ are even multiples of $\dfrac{\pi }{2}$ then $\tan x=\tan y$ implies that $x=n\pi +y$ where $n$is an integer.

Complete step by step solution: We are given a trigonometric equation $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\cos \left( \dfrac{\pi }{4}\tan \theta \right)$ and we have to determine the value of $\theta $.
We will first write both side of the equation in terms of the trigonometric functions sin.
As we know to convert sin into cos we use the formula $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $. So the equation will be,
$\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$
We will now compare both side of the trigonometric equation. So,
$\dfrac{\pi }{4}\cot \theta =\left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$
We will now simplify the equation.
$\begin{align}
  & \dfrac{\pi }{4}\cot \theta =\dfrac{\pi }{2}-\dfrac{\pi }{4}\tan \theta \\
 & \dfrac{\pi }{4}\cot \theta +\dfrac{\pi }{4}\tan \theta =\dfrac{\pi }{2} \\
 & \dfrac{\pi }{4}\left( \cot \theta +\tan \theta \right)=\dfrac{\pi }{2} \\
 & \cot \theta +\tan \theta =2
\end{align}$
We will now write the equation in terms of the trigonometric function tan by using the formula$\cot \theta =\dfrac{1}{\tan \theta }$ and simplify.
\[\begin{align}
  & \cot \theta +\tan \theta =2 \\
 & \dfrac{1}{\tan \theta }+\tan \theta =2 \\
 & \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\
 & 1+{{\tan }^{2}}\theta =2\tan \theta \\
 & 1+{{\tan }^{2}}\theta -2\tan \theta =0
\end{align}\]
Using expansion formula ${{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}$in the equation above.
\[\begin{align}
  & 1+{{\tan }^{2}}\theta -2\tan \theta =0 \\
 & {{\left( \tan \theta -1 \right)}^{2}}=0 \\
 & \tan \theta -1=0 \\
 & \tan \theta =1
\end{align}\]
We know that $\tan \dfrac{\pi }{4}=1$. So,
\[\tan \theta =\tan \dfrac{\pi }{4}\]
Applying the theorem here we will get,
$\theta =n\pi +\dfrac{\pi }{4}$, where $n\in Z$.
The value of $\theta $ for the trigonometric equation $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ is $\theta =n\pi +\dfrac{\pi }{4}$

Option ‘A’ is correct

Note: We have used the principal solution of sin after writing the equation in terms of sin that is here $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$ because we have to further reduce this equation as it has trigonometric function $\cot ,\tan $in it.
Instead of writing the equation $\cot \theta +\tan \theta =2$ in terms of tan we could have also used function sin and cos to simplify the equation and using theorems accordingly.