Question

Find the principal value of ${{\sec }^{-1}}\left( -{2} \right)$.

Answer 1

For the above question we have to know about the principal values of an inverse trigonometric function. Principal value of an inverse trigonometric function is a value that belongs to the principal branch of the range of the function. We know that the principal branch of range of ${{\sec }^{-1}}x$ is \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.

Complete step-by-step solution:
We have been given the expression ${{\sec }^{-1}}\left( - {2} \right)$.
Now we know that the principal value means the value which lies between the defined range of the function.
We know that for the inverse trigonometric function ${{\sec }^{-1}}x$ the range is \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
We also know that the value of ${{\sec}}\left( \dfrac{2\pi }{3} \right)=-{2}$.
So by substituting the value of $\left( -{2} \right)$ in the given expression we get as follows:
${{\sec }^{-1}}\left( -{2} \right)={{\sec }^{-1}}\left[ \sec \left( \dfrac{2\pi }{3} \right) \right]$
Now we must know that ${{\sec }^{-1}}\sec \theta =\theta$ where ‘$\theta$’ must lie between \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
And we know that the angle $\dfrac{2\pi }{3}\in \left[ 0,\pi \right]$
Therefore, we get that the angle lies in the range of the given function. So, now we can say that the principal value of the given function is
${{\sec }^{-1}}\left( -{2} \right)=\dfrac{2\pi }{3}$
Therefore, the principal value of ${{\sec }^{-1}}\left( -{2} \right)$ is $\dfrac{2\pi }{3}$.

Note: Be careful while finding the principal value of inverse trigonometric functions and do check it once that the value must lie between the principal branch of the range of the function. Don’t get confused that the value of ${{\sec }^{-1}}\left( -{2} \right)=\dfrac{\pi }{3}$ which is wrong as we know that $\sec \dfrac{\pi }{3}= {2}$. So be careful at this point not to make any such silly mistakes.