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Question

Answers

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) $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$

Answer
Verified

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 $\theta =n\pi +\left( \dfrac{\pi }{4} \right)$