How do you evaluate $\sec 23\deg 2^{'}?$

Hint: As we know that above question includes trigonometry as $\sec $ or secant is a trigonometric ratio. In this question we will use the reciprocal relation between each pair of the trigonometric function i.e. we know that $\sec \theta $ can be written in the inverse form of $\dfrac{1}{{\cos \theta }}$. With the help of the reciprocal function we will solve this question.

Complete step by step solution:
As per the question we have $\sec 23\deg 2^{'}.$
Let us call $x = {23^ \circ }2^{'}$. First we will convert the full in form of degrees. We will convert $2$minutes into degrees by dividing with $60.$ It can be written as $\dfrac{2}{{60}} = 0.033$.
Now add both the terms; ${23^ \circ } + {0.033^ \circ } = {23.0.33^ \circ }$.
We can write $\sec x = \dfrac{1}{{\cos x}}$, so we have to find the value of $\cos 23.033$.
Now the value of $\cos 23.033$ is $0.9205$. By putting the value we have $\sec x = \dfrac{1}{{0.92}} = 1.09$.
Hence the required answer is $1.09$.

Note: We should know that the value of $\cos 23$ is the same as in radians. To obtain the value in radian we have to multiply $23$ by $\dfrac{\pi }{{180}}$. So we can write the value as $\cos 23 = \cos \left( {\dfrac{{23}}{{180}} \times \pi } \right)$. We should also know that the angle is between $0$ and $90$, so it lies in the first quadrant and the value for sin, cos and tan are positive in this quadrant.