Question

# If a trigonometric equation is given as $\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$, then $\theta$= A) $n\pi +\left( \dfrac{\pi }{2} \right)$B) $n\pi +\left( \dfrac{\pi }{3} \right)$C) $n\pi +\left( \dfrac{\pi }{4} \right)$D) $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$

Hint: We solve this problem by converting either sin to cos (or) cos to sin using the formula $\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$ and then we equate both the angles. After changing the angles into the same trigonometric identity, we equate the angles and we solve them to find the value of $\theta$.

We use the formula,
$\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$
Let us consider the given trigonometric equation
$\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$
Now, we convert cos into sin to equate the angles. So, we get,
$\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{4}.\tan \theta \right)$
As the trigonometric terms on both sides are equal. By equating both the angles, we get
\begin{align} & \Rightarrow \dfrac{\pi }{4}.\cot \theta =\dfrac{\pi }{2}-\dfrac{\pi }{4}.\tan \theta \\ & \Rightarrow \dfrac{\pi }{4}.\cot \theta +\dfrac{\pi }{4}.\tan \theta =\dfrac{\pi }{2} \\ & \Rightarrow \dfrac{\pi }{4}\left( \cot \theta +\tan \theta \right)=\dfrac{\pi }{2} \\ \end{align}
Now we convert $\cot \theta$ to $\tan \theta$ , we get
$\dfrac{\pi }{4}\left( \dfrac{1}{\tan \theta }+\tan \theta \right)=\dfrac{\pi }{2}$
After cancelling $\dfrac{\pi }{4}$ on both the sides, we get
\begin{align} & \Rightarrow \dfrac{1}{\tan \theta }+\tan \theta =2 \\ & \Rightarrow \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\ & \Rightarrow 1+{{\tan }^{2}}\theta =2\tan \theta \\ & \Rightarrow {{\tan }^{2}}\theta -2\tan \theta +1=0 \\ \end{align}
Using the formula of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we can factorize the above quadratic equation as,
\begin{align} & \Rightarrow {{\left( \tan \theta -1 \right)}^{2}}=0 \\ & \Rightarrow \tan \theta -1=0 \\ & \Rightarrow \tan \theta =1 \\ & \Rightarrow \tan \theta =\tan \dfrac{\pi }{4} \\ \end{align}
But we have to find the general solution for $\tan \theta =1$.
Let us consider the formula for finding the general solution.
If $\alpha$ is the principal solution of $\tan \theta =k$, then the general solution of $\tan \theta$ is $n\pi +\alpha$.
Using the above formula we can write the general solution for $\tan \theta =1$ as,
The general solution for $\tan \theta =\tan \dfrac{\pi }{4}$ is $n\pi +\left( \dfrac{\pi }{4} \right)$
Therefore, if $\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$, then $\theta =n\pi +\left( \dfrac{\pi }{4} \right)$

So, the correct answer is “Option C”.

Note: One can make a mistake by considering the general of solution of $\tan \theta =1$ as $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ instead of $n\pi +\left( \dfrac{\pi }{4} \right)$. So, one need to be careful while writing the general terms for trigonometric identities. For example, if we consider $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ as the general solution for $\tan \theta =1$, then by taking n=1, $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ gives us -1, which is not the solution for $\tan \theta =1$. So, one must be careful while writing the general solution for the trigonometric identities.