Question

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

Answer 1

Hint: To find the principal value of ${{\sec }^{-1}}(2)$, we need to find the value of x for which sec(x) = 2. Further, for the condition of it being a principal value, this solution of x must lie within the following range given by $[0,\pi ]-\left\{ \dfrac{\pi }{2} \right\}$.

Complete step-by-step answer:
For finding the principal value of ${{\sec }^{-1}}(2)$, we have to find the solution for the trigonometric equation sec(x) = 2. However, since trigonometric angles are periodic in nature, there are multiple solutions (which will repeat at regular intervals). However, since we have to find the principal value, we have to find the solution within the given interval (suitable for principal value of ${{\sec }^{-1}}x$). This range is given by $[0,\pi ]-\left\{ \dfrac{\pi }{2} \right\}$. Now, we solve the equation sec(x) = 2. Thus, we have,
sec(x) = 2
cos x = $\dfrac{1}{2}$ (since, sec(x) = $\dfrac{1}{\cos x}$)
cos x = cos 60
x = $2n\pi \pm \dfrac{\pi }{3}$ (Since, 60 degrees in radians is $\dfrac{\pi }{3}$)
Now, we can see that only one solution lies within the range $[0,\pi ]-\left\{ \dfrac{\pi }{2} \right\}$. This is clearly $\dfrac{\pi }{3}$. Hence, the principal value of ${{\sec }^{-1}}(2)$ is $\dfrac{\pi }{3}$.

Note: Solving questions related to inverse trigonometric equations (like the one in the problem above), these are similar to solving the normal trigonometric equations. The only difference is that in addition, we need to take care of the additional constraint of principal value (that is the solutions of the inverse trigonometric equation must lie within the required range). These ranges are different for different inverse trigonometric equations (like ${{\sin }^{-1}}x,{{\cos }^{-1}}x$ and so on).