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# Find the value of ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$

Last updated date: 20th Jun 2024
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Hint: We know the range of both the inverse trigonometric functions and also, $\sqrt{3}$ and $-\sqrt{3}$ are basic values for tan and cot inverse functions, i.e. , they are the kind of values whose values are known when they are put in the cot and tan inverse functions. Thus, we can find their values very easily. So we will find these individual values in the range of these functions and then subtract them. This will give us our answer.

Now, we know that the range of ${{\tan }^{-1}}x$ is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
We also know that in the provided range the value of ${{\tan }^{-1}}\sqrt{3}=\dfrac{\pi }{3}$
Now, we also know that the range of ${{\cot }^{-1}}x$ is $\left( 0,\pi \right)$
We also know that in the provided range the value of ${{\cot }^{-1}}\left( -\sqrt{3} \right)=\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}$
Now since we know the value of both ${{\tan }^{-1}}\left( \sqrt{3} \right)$ and ${{\cot }^{-1}}\left( -\sqrt{3} \right)$ we can subtract these two values and obtain our required values.
Thus, the value of ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$is given as:
\begin{align} & \Rightarrow \dfrac{\pi }{3}-\dfrac{5\pi }{6} \\ & \Rightarrow -\dfrac{3\pi }{6} \\ & \Rightarrow -\dfrac{\pi }{2} \\ \end{align}

Thus, the value of ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$ is $-\dfrac{\pi }{2}$

Note: This question can also be done in the following way:
We know that the value of ${{\cot }^{-1}}\left( -x \right)$ is given as $\pi -{{\cot }^{-1}}x$
So we can write the value of ${{\cot }^{-1}}\left( -\sqrt{3} \right)$ in the same way.
Thus, the value of ${{\cot }^{-1}}\left( -\sqrt{3} \right)=\pi -{{\cot }^{-1}}\left( \sqrt{3} \right)$
So, the value of ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$becomes:
\begin{align} & \Rightarrow {{\tan }^{-1}}\sqrt{3}-\left( \pi -{{\cot }^{-1}}\left( \sqrt{3} \right) \right) \\ & \Rightarrow {{\tan }^{-1}}\sqrt{3}-\pi +{{\cot }^{-1}}\left( \sqrt{3} \right) \\ & \Rightarrow {{\tan }^{-1}}\sqrt{3}+{{\cot }^{-1}}\left( \sqrt{3} \right)-\pi \\ \end{align}
Now, we know that the value of ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ for all $x\in \mathbb{R}$
Thus the value of ${{\tan }^{-1}}\sqrt{3}+{{\cot }^{-1}}\left( \sqrt{3} \right)-\pi$ is given as:
\begin{align} & \Rightarrow \dfrac{\pi }{2}-\pi \\ & \Rightarrow -\dfrac{\pi }{2} \\ \end{align}