Question

How do you find the exact value of $\sec \dfrac{\pi }{2}$?

Answer 1

Hint: In the above problem, we are asked to find the exact value of $\sec \dfrac{\pi }{2}$. We know that secant is the reciprocal of cosine so we can write this secant expression as reciprocal of cosine. So, the given expression is the secant of $\dfrac{\pi }{2}$ and writing this expression as the reciprocal of cosine of $\dfrac{\pi }{2}$. And then substitute the value of $\cos \dfrac{\pi }{2}$.

Complete step by step solution:
The trigonometric expression given in the above problem of which we are asked to find the exact value:
$\sec \dfrac{\pi }{2}$
Now, we know that secant is the reciprocal of cosine so writing the $\sec \dfrac{\pi }{2}$ as the reciprocal of $\cos \dfrac{\pi }{2}$. And reciprocal of $\cos \dfrac{\pi }{2}$ means $\sec \dfrac{\pi }{2}$ is equal to:
$\dfrac{1}{\cos \dfrac{\pi }{2}}$
Now, according to the trigonometric ratio values we know that the value of $\cos \dfrac{\pi }{2}$ is equal to 0 so substituting this value of $\cos \dfrac{\pi }{2}$ as 0 in the above expression we get,
$\dfrac{1}{0}$
And we know that if we have $\dfrac{1}{0}$ form then this is not defined or infinity.
From the above, we got the exact value of $\sec \dfrac{\pi }{2}$ as not defined.

Note: The alternate approach to the above problem is that if you remember the value of $\sec \dfrac{\pi }{2}$ in the same way you did for $\cos \dfrac{\pi }{2}$ then the above problem has a one line solution. In case you don’t remember the value then some angles of sine and cosine values we always know and then reciprocal the value of $\cos \dfrac{\pi }{2}$ which we have shown in the above solution.
The point to be noted that if you could not remember the value of $\sec \dfrac{\pi }{2}$ or the other values of secant like $\sec \dfrac{\pi }{6},\sec \dfrac{\pi }{4},\sec \dfrac{\pi }{3}$ then there is a way which we can find the solution by taking the reciprocal of cosine but if you don’t remember the value of $\cos \dfrac{\pi }{6},\cos \dfrac{\pi }{3},\cos \dfrac{\pi }{4},\cos \dfrac{\pi }{2},\cos 0$ then you cannot solve this problem.