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Question: Evaluate the following expression \[{{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^...

Evaluate the following expression tan⁡−1(−13)+cot⁡−1(13)+tan⁡−1(sin⁡(−π2)){{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( \sin \left( \dfrac{-\pi }{2} \right) \right).

Explanation

Solution

Hint:In order to solve this question, we should know a few identities like, sin⁡(−θ)=−sin⁡θ,tan⁡−1(−x)=−tan⁡−1x\sin \left( -\theta \right)=-\sin \theta ,{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x. We should also know about a few trigonometric ratios like, tan⁡π6=13,cot⁡π3=13\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}},\cot \dfrac{\pi }{3}=\dfrac{1}{\sqrt{3}} and tan⁡π4=1\tan \dfrac{\pi }{4}=1. By using these we can solve this question.

Complete step-by-step answer:
In this question we have been given an expression, that is, tan⁡−1(−13)+cot⁡−1(13)+tan⁡−1(sin⁡(−π2)){{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( \sin \left( \dfrac{-\pi }{2} \right) \right) and we have been asked to simplify it. To solve this, we will start from sin⁡(−π2)\sin \left( \dfrac{-\pi }{2} \right). We know that sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta , so we can write sin⁡(−π2)\sin \left( \dfrac{-\pi }{2} \right) as −(sin⁡π2)-\left( \sin \dfrac{\pi }{2} \right). Therefore, we will get the expression as,
tan⁡−1(−13)+cot⁡−1(13)+tan⁡−1(−sin⁡π2){{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( -\sin \dfrac{\pi }{2} \right)
We also know that, sin⁡π2=1\sin \dfrac{\pi }{2}=1. So, we can write −sin⁡π2=−1-\sin \dfrac{\pi }{2}=-1. Hence, by substituting this value in the above expression, we will get,
tan⁡−1(−13)+cot⁡−1(13)+tan⁡−1(−1){{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( -1 \right)
Now, we know that tan⁡−1(−x)=−tan⁡−1x{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x. So, we can write the given expression as,
−tan⁡−1(13)+cot⁡−1(13)−tan⁡−1(1)-{{\tan }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\tan }^{^{-1}}}\left( 1 \right)
Now, we know that tan⁡π6=13,cot⁡π3=13\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}},\cot \dfrac{\pi }{3}=\dfrac{1}{\sqrt{3}} and tan⁡π4=1\tan \dfrac{\pi }{4}=1. So, we can write them as, π6=tan⁡−113,π3=cot⁡−113\dfrac{\pi }{6}={{\tan }^{-1}}\dfrac{1}{\sqrt{3}},\dfrac{\pi }{3}={{\cot }^{-1}}\dfrac{1}{\sqrt{3}} and π4=tan⁡−11\dfrac{\pi }{4}={{\tan }^{-1}}1. Hence, we get the expression as,
−π6+π3−π4\dfrac{-\pi }{6}+\dfrac{\pi }{3}-\dfrac{\pi }{4}
Now, we will take the LCM of the above terms. So, we will get,
−2π+4π−3π12\dfrac{-2\pi +4\pi -3\pi }{12}
Now, we know that the like terms show arithmetic operation, so we can write the expression as follows,

& \dfrac{4\pi -5\pi }{12} \\\ & \Rightarrow \dfrac{-\pi }{12} \\\ \end{aligned}$$ Hence, we can say that the value of the expression, that is, $${{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( \sin \left( \dfrac{-\pi }{2} \right) \right)$$ is $$\dfrac{-\pi }{12}$$. Note: While solving this question, the possible mistake that the students can make is by ignoring the negative sign of $\left( \dfrac{-\pi }{2} \right)$ and ${{\tan }^{-1}}\left( \dfrac{-1}{\sqrt{3}} \right)$ in a hurry and this will give them the wrong answer. So, the students must be careful while solving this question. Also, the students must remember the basic trigonometric ratios like, $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}},\cot \dfrac{\pi }{3}=\dfrac{1}{\sqrt{3}}$ and $\tan \dfrac{\pi }{4}=1$.