Question

The value of $\sec {{90}^{o}}$ is equal to:
(A). 1
(B). $\dfrac{2}{\sqrt{3}}$
(C). $\sqrt{2}$
(D). None of these

Answer 1

Hint: The given problem is related to trigonometric ratios. Use the relation between cosine and secant of an angle, then use the value of \[\cos {{90}^{o}}=0\] to find the value of $\sec {{90}^{o}}$ .

Complete step-by-step solution -
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $
Now, from the graph of cosine function, we can see that the value of $\cos {{90}^{o}}$ is equal to 0.
Now, we know that the relation between the cosine and the secant of an angle (say $\theta $ ) is given as $\sec \theta =\dfrac{1}{\cos \theta }$ . So, we can say that the value of $\sec {{90}^{o}}$ is given as $\sec {{90}^{o}}=\dfrac{1}{\cos {{90}^{o}}}$ .
But we know that the value of $\cos {{90}^{o}}$ is equal to 0.
$\Rightarrow \sec {{90}^{o}}=\dfrac{1}{0}$
We know that, when any number (except 0) when divided by 0 gives an infinite value. So, we can say that the value of $\dfrac{1}{0}$ will be equal to infinity.
$\Rightarrow \sec {{90}^{o}}=\infty $
On checking the options, we can see that none of the first three options has the value $\infty $ . Hence, option D is the correct option.

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two.