Question

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

Answer 1

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

Complete step-by-step answer:
In this question we have been given an expression, that is, \[{{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( \sin \left( \dfrac{-\pi }{2} \right) \right)\] and we have been asked to simplify it. To solve this, we will start from $\sin \left( \dfrac{-\pi }{2} \right)$. We know that $\sin \left( -\theta \right)=-\sin \theta $, so we can write $\sin \left( \dfrac{-\pi }{2} \right)$ as $-\left( \sin \dfrac{\pi }{2} \right)$. Therefore, we will get the expression as,
\[{{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( -\sin \dfrac{\pi }{2} \right)\]
We also know that, $\sin \dfrac{\pi }{2}=1$. So, we can write $-\sin \dfrac{\pi }{2}=-1$. Hence, by substituting this value in the above expression, we will get,
\[{{\tan }^{^{-1}}}\left( \dfrac{-1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\tan }^{^{-1}}}\left( -1 \right)\]
Now, we know that ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x$. So, we can write the given expression as,
\[-{{\tan }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)+{{\cot }^{^{-1}}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\tan }^{^{-1}}}\left( 1 \right)\]
Now, we know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}},\cot \dfrac{\pi }{3}=\dfrac{1}{\sqrt{3}}$ and $\tan \dfrac{\pi }{4}=1$. So, we can write them as, $\dfrac{\pi }{6}={{\tan }^{-1}}\dfrac{1}{\sqrt{3}},\dfrac{\pi }{3}={{\cot }^{-1}}\dfrac{1}{\sqrt{3}}$ and $\dfrac{\pi }{4}={{\tan }^{-1}}1$. Hence, we get the expression as,
$\dfrac{-\pi }{6}+\dfrac{\pi }{3}-\dfrac{\pi }{4}$
Now, we will take the LCM of the above terms. So, we will get,
$\dfrac{-2\pi +4\pi -3\pi }{12}$
Now, we know that the like terms show arithmetic operation, so we can write the expression as follows,
\[\begin{align}
  & \dfrac{4\pi -5\pi }{12} \\
 & \Rightarrow \dfrac{-\pi }{12} \\
\end{align}\]
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 $\tan \dfrac{\pi }{4}=1$.