Question

Is sec the inverse of cos?

Answer 1

Hint: Now we know that cos is nothing but the ratio of adjacent side and hypotenuse. Similarly sec is nothing but the ratio of hypotenuse and adjacent side. Now if ${{f}^{-1}}$ is an inverse function of $f$ and $f\left( x \right)=y$ then we have ${{f}^{-1}}\left( y \right)=x$ . Hence we will check sec is the inverse function of cos.

Complete step by step solution:
Now first let us understand the concept of functions and inverse functions.
Now a function can be imagined as a machine which takes an input to a unique output.
Hence let us say we have a value x. then the function takes the value and given an output y. Hence we have $f\left( x \right)=y$ .
Now let us understand the concept of inverse function. Inverse function of a function is the function which reverses the role of the actual function.
Hence if we have f as a function where $f\left( x \right)=y$ . then the inverse of $f$ is denoted by ${{f}^{-1}}$ and we have ${{f}^{-1}}\left( y \right)=x$ .
Now let us understand cos function. $\cos $is a trigonometric ratio which gives the ratio of adjacent side and hypotenuse. Now hence $\cos \theta $ is a function which takes the value of angle $\theta $ and then gives the corresponding cos value in the triangle.
Now similarly we have the trigonometric ratio $\sec $ . $\sec $ is a trigonometric ratio defined as the ratio of hypotenuse and adjacent side. Hence we have $\sec \theta =\dfrac{1}{\cos \theta }$ . Now note that sec is the inverse of cos but not the inverse function of cos. As if we have $\cos 0=1$ then $\sec 1\ne 0$ . Hence the function sec is not the inverse function of cos.

Note: Now note that though the representation of an inverse function is ${{f}^{-1}}\left( x \right)$ we have $\dfrac{1}{f\left( x \right)}\ne {{f}^{-1}}\left( x \right)$ . Hence not to be confused between the two terms. Also note that the inverse of all functions does not exist; the function needs to satisfy certain criteria for the inverse function to exist. The criteria is the function needs to be one-one and onto.