Question

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

Answer 1

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 $x=\dfrac{\pi }{3}$.

Complete step-by-step answer:
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

& \cos y=\dfrac{1}{\sec y} \\\ & =-\dfrac{1}{2} \end{aligned}$$ We have to now find the value of $$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{aligned} & y=\pi -\dfrac{\pi }{3} \\\ & =\dfrac{3\pi -\pi }{3} \\\ & =\dfrac{2\pi }{3} \end{aligned}$$ Thus we have $$\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{aligned} & x-y=\dfrac{\pi }{3}-\dfrac{2\pi }{3} \\\ & =-\dfrac{\pi }{3} \end{aligned}$$ Therefore the principal value of $${{\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.