Question

The principal value of \[{{\tan }^{-1}}\left( \cot \dfrac{43\pi }{4} \right)\]
(a) \[\dfrac{-3\pi }{4}\]
(b) \[\dfrac{3\pi }{4}\]
(c) \[\dfrac{-\pi }{4}\]
(d) \[\dfrac{\pi }{4}\]

Answer 1

Hint: We solve this problem by converting the given angle that is \[\dfrac{43\pi }{4}\] into \[\left( n\pi +\theta \right)\] where \[\theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\] because the range of inverse tangent ratio is \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\] so that the principal value lies in that range only.
We have the standard formula of cotangent trigonometric ratio that is
\[\cot \left( n\pi +\theta \right)=\cot \theta ,\forall n\in \mathbb{N}\]
We also have the conversion of cotangent to tangent that is
\[\Rightarrow \cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)\]
Then we have the direct formula of inverse tangent ratio that is
\[\Rightarrow {{\tan }^{-1}}\left( \tan \theta \right)=\theta ,\forall \theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]

Complete step by step answer:
We are asked to find the principal value of \[{{\tan }^{-1}}\left( \cot \dfrac{43\pi }{4} \right)\]
Let us assume that the required value as
\[\Rightarrow A={{\tan }^{-1}}\left( \cot \dfrac{43\pi }{4} \right)\]
Now, let us try to convert the angle \[\dfrac{43\pi }{4}\] into \[\left( n\pi +\theta \right)\]
Here we can see that the number in the numerator 40 can be written as \[40+3\] because we can get in the required form.
So, by converting the numerator in the given expression we get
\[\begin{align}
  & \Rightarrow A={{\tan }^{-1}}\left( \cot \left( \dfrac{40\pi +3\pi }{4} \right) \right) \\
 & \Rightarrow A={{\tan }^{-1}}\left( \cot \left( 10\pi +\dfrac{3\pi }{4} \right) \right) \\
\end{align}\]
We know that the standard formula of cotangent trigonometric ratio that is
\[\cot \left( n\pi +\theta \right)=\cot \theta ,\forall n\in \mathbb{N}\]
By using this formula in above equation we get
\[\Rightarrow A={{\tan }^{-1}}\left( \cot \dfrac{3\pi }{4} \right)\]
Now, let us convert the cotangent ratio to tangent ratio.
We know that the conversion of cotangent to tangent that is
\[\Rightarrow \cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)\]
By using this conversion in above equation we get
\[\begin{align}
  & \Rightarrow A={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{2}-\dfrac{3\pi }{4} \right) \right) \\
 & \Rightarrow A={{\tan }^{-1}}\left( \tan \left( \dfrac{2\pi -3\pi }{4} \right) \right) \\
 & \Rightarrow A={{\tan }^{-1}}\left( \tan \left( \dfrac{-\pi }{4} \right) \right) \\
\end{align}\]
We know that the direct formula of inverse tangent ratio that is
\[\Rightarrow {{\tan }^{-1}}\left( \tan \theta \right)=\theta ,\forall \theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]
Here we can see that the angle \[\dfrac{-\pi }{4}\] lies in the domain \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]
Therefore by applying the above formula we get
\[\Rightarrow A=\dfrac{-\pi }{4}\]
Therefore we can conclude that the principal value of given expression is \[\dfrac{-\pi }{4}\]
So, option (c) is correct answer.


Note:
 We can solve this problem in other methods.
We are asked to find the principal value of \[{{\tan }^{-1}}\left( \cot \dfrac{43\pi }{4} \right)\]
Let us assume that the required value as
\[\Rightarrow A={{\tan }^{-1}}\left( \cot \dfrac{43\pi }{4} \right)\]
Now, let us try to convert the angle \[\dfrac{43\pi }{4}\] into \[\left( n\pi -\theta \right)\]
Here we can see that the number in the numerator 40 can be written as \[44-1\] because we can get in the required form.
So, by converting the numerator in the given expression we get
\[\begin{align}
  & \Rightarrow A={{\tan }^{-1}}\left( \cot \left( \dfrac{44\pi -\pi }{4} \right) \right) \\
 & \Rightarrow A={{\tan }^{-1}}\left( \cot \left( 11\pi -\dfrac{\pi }{4} \right) \right) \\
\end{align}\]
We know that the standard formula of cotangent trigonometric ratio that is
\[\cot \left( n\pi -\theta \right)=-\cot \theta ,\forall n\in \mathbb{N}\]
By using this formula in above equation we get
\[\Rightarrow A={{\tan }^{-1}}\left( -\cot \dfrac{\pi }{4} \right)\]
We know that the standard formula of inverse tangent ratio that is
\[{{\tan }^{-1}}\left( -\theta \right)=-{{\tan }^{-1}}\theta \]
By using this formula we get
\[\Rightarrow A=-{{\tan }^{-1}}\left( \cot \dfrac{\pi }{4} \right)\]
Now, let us convert the cotangent ratio to tangent ratio.
We know that the conversion of cotangent to tangent that is
\[\Rightarrow \cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)\]
By using this conversion in above equation we get
\[\begin{align}
  & \Rightarrow A=-{{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{2}-\dfrac{\pi }{4} \right) \right) \\
 & \Rightarrow A=-{{\tan }^{-1}}\left( \tan \left( \dfrac{2\pi -\pi }{4} \right) \right) \\
 & \Rightarrow A=-{{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4} \right) \right) \\
\end{align}\]
We know that the direct formula of inverse tangent ratio that is
\[\Rightarrow {{\tan }^{-1}}\left( \tan \theta \right)=\theta ,\forall \theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]
Here we can see that the angle \[\dfrac{\pi }{4}\] lies in the domain \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]
Therefore by applying the above formula we get
\[\Rightarrow A=\dfrac{-\pi }{4}\]
Therefore we can conclude that the principal value of given expression is \[\dfrac{-\pi }{4}\]
So, option (c) is correct answer.