Question

${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$ is equal to
(a). $\pi$
(b). $-\dfrac{\pi }{2}$
(c). 0
(d). $2\sqrt{3}$

Answer 1

Hint: At first, find the value of terms given in the expression separately. By using fact that $\tan \dfrac{\pi }{3}$ is $\sqrt{3}$ and $\cot \left( \dfrac{5\pi }{6} \right)$ is $-\sqrt{3}$. Solve and find the value of what is asked.

Complete step by step answer:
In the question, we are given an expression ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$ and we have to find the value of expression and tell which is the correct option.

Let's find the value of ${{\tan }^{-1}}\sqrt{3}$ first. Suppose the value of ${{\tan }^{-1}}\sqrt{3}$ be ${{\theta }_{1}}$ . Then, we can say that $\tan {{\theta }_{1}}$ is equal to $\sqrt{3}$ . We know that ${{\tan }^{-1}}\dfrac{\pi }{3}$ is $\sqrt{3}$. So, we can say that $\tan {{\theta }_{1}}$ is equal $\tan \dfrac{\pi }{3}$ or ${{\theta }_{1}}$ is equal to $\dfrac{\pi }{3}$ .
Now let’s take or suppose the value of ${{\cot }^{-1}}\left( -\sqrt{3} \right)$ be ${{\theta }_{2}}$. So, we can say that $\cot {{\theta }_{2}}=-\sqrt{3}$ . So, we can say that $\cot {{\theta }_{2}}$ is equal to $\cot \left( \dfrac{5\pi }{6} \right)$ or ${{\theta }_{2}}$ is equal to $\dfrac{5\pi }{6}$ .
Now, we have to find the value of ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$ or ${{\theta }_{1}}-{{\theta }_{2}}$ .
As we know that ${{\theta }_{1}}$ is $\dfrac{\pi }{3}$ and ${{\theta }_{2}}$ is $\dfrac{5\pi }{6}$ .
So, we can write as $\left( \dfrac{\pi }{3}-\dfrac{5\pi }{6} \right)$ or $\left( \dfrac{2\pi -5\pi }{6} \right)$ or $-\dfrac{\pi }{2}$ .
Hence the correct option is ‘B’.

Note: We can also do it using another method. We can write ${{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}\left( -\sqrt{3} \right)$ as ${{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ . Then we will apply identity ${{\tan }^{-1}}x-{{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$ where we will take x as $\sqrt{3}$ and y as $\dfrac{-1}{\sqrt{3}}$ .