Question

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

Answer 1

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 \[x=\dfrac{\pi }{3}\].

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
\[\begin{align}
  & \cos y=\dfrac{1}{\sec y} \\
 & =-\dfrac{1}{2}
\end{align}\]
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.