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# Determine the principal value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$.

Hint: In this question, We will first assume that ${{\tan }^{-1}}\sqrt{3}=x$, which implies that $\sqrt{3}=\tan x$. Now we have to find the value of $x$ for which that value of $\tan x=\sqrt{3}$. We will get that $x=\dfrac{\pi }{3}$. Then we will assume that $y={{\sec }^{-1}}\left( -2 \right)$. This will imply $\sec y=-2$. Now we have to find the value of $y$ for which that value of $\sec y=-2$. Now since $\cos \left( \pi -\dfrac{\pi }{3} \right)=-\dfrac{1}{2}$, thus we will get $\sec \left( \pi -\dfrac{\pi }{3} \right)=-2$. So the principal value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$ can be calculated by subtracting $y=\left( \pi -\dfrac{\pi }{3} \right)$ from $x=\dfrac{\pi }{3}$.

In order to find the principal value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$, we have to first calculate the principal value of ${{\tan }^{-1}}\sqrt{3}$ and the principal value of ${{\sec }^{-1}}\left( -2 \right)$ and then add both the principal values to get the desired answer.
So let us suppose that the principal value of ${{\tan }^{-1}}\sqrt{3}$ is given by $x$.
That is let ${{\tan }^{-1}}\sqrt{3}=x$.
Thus we get $\tan x=\sqrt{3}$.
Now we have to find the value of $x$ for which the value of $\tan x=\sqrt{3}$.
Since $\tan x=\sqrt{3}=\tan \dfrac{\pi }{3}$, therefore we have $x=\dfrac{\pi }{3}$.
Also since $x=\dfrac{\pi }{3}\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, hence $x=\dfrac{\pi }{3}$ is a principal value of ${{\tan }^{-1}}\sqrt{3}$.
Now again let us suppose that the principal value of ${{\sec }^{-1}}\left( -2 \right)$ is given by $y$.
That is let $y={{\sec }^{-1}}\left( -2 \right)$.
Thus we have $\sec y=-2$.
Now we have to find the value of $y$ for which the value of $\sec y=-2$.
Now we know that $\sec y=\dfrac{1}{\cos y}$.
Therefore we get
\begin{align} & \cos y=\dfrac{1}{\sec y} \\ & =-\dfrac{1}{2} \end{align}
We have to now find the value of $y$ such that $\cos y=-\dfrac{1}{2}$.

Now since $\cos \theta =\dfrac{1}{2}=\cos \dfrac{\pi }{3}$, thus $\theta =\dfrac{\pi }{3}$.

But we have the negative value of $\cos y$. Thus $y$ must belong to the second quadrant.
Hence the value of $y$ is given by
\begin{align} & y=\pi -\dfrac{\pi }{3} \\ & =\dfrac{3\pi -\pi }{3} \\ & =\dfrac{2\pi }{3} \end{align}
Thus we have $\cos y=-\dfrac{1}{2}=\cos \dfrac{2\pi }{3}$.
Hence $y=\dfrac{2\pi }{3}$ is a principal value of ${{\sec }^{-1}}\left( -2 \right)$.
Now we will finally calculate the principal value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$.
So the principal value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$ is given by
\begin{align} & x-y=\dfrac{\pi }{3}-\dfrac{2\pi }{3} \\ & =-\dfrac{\pi }{3} \end{align}
Therefore the principal value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)$ is $-\dfrac{\pi }{3}$.

Note: In this problem, while calculating the value of $y$ for which the value of $\sec y=-2$ we have to take care of the fact that the value is negative. Hence $y$ must lie in the second quadrant. Accordingly we have to find the value of $y$ for which the $\sec y=-2$.do not just take $y=\dfrac{\pi }{3}$ because for $y=\dfrac{\pi }{3}$ the value of $\sec y$ is equals to 2.