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

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

Explanation

Solution

We know the range of both the inverse trigonometric functions and also, 3\sqrt{3} and −3-\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.

Complete step by step answer:
Now, we know that the range of tan⁡−1x{{\tan }^{-1}}x is (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)
We also know that in the provided range the value of tan⁡−13=π3{{\tan }^{-1}}\sqrt{3}=\dfrac{\pi }{3}
Now, we also know that the range of cot⁡−1x{{\cot }^{-1}}x is (0,π)\left( 0,\pi \right)
We also know that in the provided range the value of cot⁡−1(−3)=π−π6=5π6{{\cot }^{-1}}\left( -\sqrt{3} \right)=\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}
Now since we know the value of both tan⁡−1(3){{\tan }^{-1}}\left( \sqrt{3} \right) and cot⁡−1(−3){{\cot }^{-1}}\left( -\sqrt{3} \right) we can subtract these two values and obtain our required values.
Thus, the value of tan⁡−13−cot⁡−1(−3){{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)is given as:
⇒π3−5π6 ⇒−3π6 ⇒−π2 \begin{aligned} & \Rightarrow \dfrac{\pi }{3}-\dfrac{5\pi }{6} \\\ & \Rightarrow -\dfrac{3\pi }{6} \\\ & \Rightarrow -\dfrac{\pi }{2} \\\ \end{aligned}

Thus, the value of tan⁡−13−cot⁡−1(−3){{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right) is −π2-\dfrac{\pi }{2}

Note: This question can also be done in the following way:
We know that the value of cot⁡−1(−x){{\cot }^{-1}}\left( -x \right) is given as π−cot⁡−1x\pi -{{\cot }^{-1}}x
So we can write the value of cot⁡−1(−3){{\cot }^{-1}}\left( -\sqrt{3} \right) in the same way.
Thus, the value of cot⁡−1(−3)=π−cot⁡−1(3){{\cot }^{-1}}\left( -\sqrt{3} \right)=\pi -{{\cot }^{-1}}\left( \sqrt{3} \right)
So, the value of tan⁡−13−cot⁡−1(−3){{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)becomes:
⇒tan⁡−13−(π−cot⁡−1(3)) ⇒tan⁡−13−π+cot⁡−1(3) ⇒tan⁡−13+cot⁡−1(3)−π \begin{aligned} & \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{aligned}
Now, we know that the value of tan⁡−1x+cot⁡−1x=π2{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} for all x∈Rx\in \mathbb{R}
Thus the value of tan⁡−13+cot⁡−1(3)−π{{\tan }^{-1}}\sqrt{3}+{{\cot }^{-1}}\left( \sqrt{3} \right)-\pi is given as:
⇒π2−π ⇒−π2 \begin{aligned} & \Rightarrow \dfrac{\pi }{2}-\pi \\\ & \Rightarrow -\dfrac{\pi }{2} \\\ \end{aligned}