Question

The principal value of \[{{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)\] is
\[A.\dfrac{-3\pi }{4}\]
\[B.\dfrac{3\pi }{4}\]
\[C.\dfrac{-\pi }{4}\]
\[D.\dfrac{\pi }{4}\]

Answer 1

Hint: In the question, we need to find the principal value of the function \[{{\tan }^{-1}}\]. The principal value of \[{{\tan }^{-1}}(x)\] is \[\theta \] if \[\dfrac{-\pi }{2}<\theta <\dfrac{\pi }{2}\] and its general value is \[n\pi +\theta \]. To solve use basic trigonometry identities formula like \[\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta \]

Complete step-by-step answer:
From the question, we can rewrite it as

\[{{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)={{\tan }^{-1}}\left( cot\left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right) \right)\]
Now, we will apply the identity \[\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta \] in the above expression. Then, we will get
\[\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( -\tan \dfrac{\pi }{4} \right) \\
 & \because \tan \left( -\theta \right)=-\tan \theta \\
 & \Rightarrow {{\tan }^{-1}}\left( \tan \left( -\dfrac{\pi }{4} \right) \right) \\
\end{align}\]
Now, we know that \[{{\tan }^{-1}}\left( \tan \theta \right)=\theta \], so we will use it and get
\[\Rightarrow \left( -\dfrac{\pi }{4} \right)\]
Hence, the principal value of \[{{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)\] is \[\dfrac{-\pi }{4}\] . Principal value for tan function always lies between \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]. So, if somehow the obtained value is less than \[-\dfrac{\pi }{2}\] or greater than \[\dfrac{\pi }{2}\], then we must have to bring the principal value in between \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\] using trigonometric identities as explained above.

Therefore, option C is the correct one.

Note: Sometimes the question would be like finding the principal value of \[\left( {{\tan }^{-1}}\left( -\dfrac{\pi }{6} \right) \right)\]. For this question, be careful, because in this question \[\dfrac{-\pi }{6}\] is lying between \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]. Therefore, its principal value would be directly \[\dfrac{-\pi }{6}\]. If \[\dfrac{-\pi }{6}\]wouldn’t be lying between \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]. Then, \[\dfrac{-\pi }{6}\] wouldn’t be the principal value of this question. For example: for the question \[{{\tan }^{-1}}\left( \dfrac{5\pi }{3} \right)\] , principal value wouldn’t be \[\dfrac{5\pi }{3}\]. To find its principal value, we need to use the general formula \[n\pi +\theta \].
In the above question \[{{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)\], to find its principal value, in between we used identity \[\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta \]. But this formula isn't the same for \[\cot \left( \dfrac{\pi }{2}-\dfrac{\pi }{3} \right)\]. For this expression, the identity is \[\cot \left( \dfrac{\pi }{2}-x \right)\]\[=\tan x\]. So, be careful while using identity.So, for\[{{\tan }^{-1}}\left( \cot \left( \dfrac{\pi }{2}-\dfrac{\pi }{3} \right) \right)\] ,the principal value would be \[\dfrac{\pi }{3}\] .