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Question: Find the value of \({\tan ^{ - 1}}\sqrt 3 - {\cot ^{ - 1}}( - \sqrt 3 )\)...

Find the value of tan⁡−13−cot⁡−1(−3){\tan ^{ - 1}}\sqrt 3 - {\cot ^{ - 1}}( - \sqrt 3 )

Explanation

Solution

Find the principal value of the two trigonometric functions and find the value and perform action. First the value of the tan⁡−13{\tan ^{ - 1}}\sqrt 3 , with the help of the principal value of tan⁡−1{\tan ^{ - 1}} and from that we will find the value of the tan⁡−1{\tan ^{ - 1}} and similar for finding the value of cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 ) by the help of its principle value and then we get the value of it and then we are going to substitute them into the main expression and get the final value which is required in the question.

Complete step by step solution:
First let's find the value of tan⁡−13{\tan ^{ - 1}}\sqrt 3 .

⇒y=tan⁡−13 ⇒tan⁡ y=3 \Rightarrow y = {\tan ^{ - 1}}\sqrt 3 \\\ \Rightarrow \tan \,y = \sqrt 3 \\\

Since3\sqrt 3 is positive, so the range of tan⁡−1{\tan ^{ - 1}} is (−π2,π2)\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)
⇒tan⁡ y=tan⁡(π3)\Rightarrow \tan \,y = \tan \left( {\dfrac{\pi }{3}} \right)
Hence, principal value of tan⁡−13{\tan ^{ - 1}}\sqrt 3 is π3\dfrac{\pi }{3}
The same way, are going to find the value of cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 )
Let us consider the given inverse term as equal to xx
⇒x=cot⁡−1(−3)\Rightarrow x = {\cot ^{ - 1}}( - \sqrt 3 )
Now we are going to find the value
⇒cot⁡x=−3\Rightarrow \cot x = - \sqrt 3
Since −3- \sqrt 3 is negative
The principal value of cot⁡−1{\cot ^{ - 1}} isπ−θ\pi - \theta
It lies between the range of (0,π)(0,\pi )
So, on substituting, we will get
⇒π−π6=5π6\Rightarrow \pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}
The value of cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 ) is 5π6\dfrac{{5\pi }}{6}.
Since, we have found the value of all the inverse trigonometric functions, we can now substitute them into the given expression from the question.
⇒tan⁡−13−cot⁡−1(−3)\Rightarrow {\tan ^{ - 1}}\sqrt 3 - {\cot ^{ - 1}}( - \sqrt 3 )
On substituting, we get

⇒π3−5π6 ⇒2π−5π6 ⇒−3π6 ⇒−π2 \Rightarrow \dfrac{\pi }{3} - \dfrac{{5\pi }}{6} \\\ \Rightarrow \dfrac{{2\pi - 5\pi }}{6} \\\ \Rightarrow \dfrac{{ - 3\pi }}{6} \\\ \Rightarrow \dfrac{{ - \pi }}{2} \\\

The above value is the final resultant solution required questions.

Thus the answer is −π2\dfrac{{ - \pi }}{2}

Note: To solve this solution, we have to be familiar with the principal values of all the inverse trigonometric functions and what range they lie in and based on the nature of the value of inverse trigonometry, we get the values.