Question

Is it true that ${\cos ^{ - 1}}x = \dfrac{1}{{\cos x}}$?

Answer 1

Hint : We usually denote the inverse of by , which is equal to $\dfrac{1}{a}$. But this is not the case with trigonometric functions. In trigonometric functions, we have to understand the difference between ${\left( {\sin x} \right)^{ - 1}}$ and ${\sin ^{ - 1}}x$. We will first see what both the expressions are actually equal to and then decide whether ${\cos ^{ - 1}}x = \dfrac{1}{{\cos x}}$ or not.

Complete step-by-step answer :
We need to check whether ${\cos ^{ - 1}}x = \dfrac{1}{{\cos x}}$ or not.
We very well know that
$\dfrac{1}{{\cos x}} = \sec x$
We see that this expression $\sec x = \dfrac{1}{{\cos x}}$ can be written as
$\sec x = \dfrac{1}{{\cos x}} = {\left( {\cos x} \right)^{ - 1}}$
So, our right hand side is equal to ${\left( {\cos x} \right)^{ - 1}}$.
Now, we need to check whether ${\cos ^{ - 1}}x = {\left( {\cos x} \right)^{ - 1}}$ or not.
We know, ${\cos ^{ - 1}}x$ is itself an inverse trigonometric function. But ${\left( {\cos x} \right)^{ - 1}}$ is the reciprocal of a trigonometric function.
Since both ${\cos ^{ - 1}}x$ and ${\left( {\cos x} \right)^{ - 1}}$ play different roles, they cannot be equal.
Hence, we conclude that ${\cos ^{ - 1}}x \ne {\left( {\cos x} \right)^{ - 1}}$
Which means ${\cos ^{ - 1}}x \ne \dfrac{1}{{\cos x}}$.
So, it is not true that ${\cos ^{ - 1}}x = \dfrac{1}{{\cos x}}$.

Note : We need to keep in mind that when we are given positive powers, then only we can say that ${\cos ^n}x = {\left( {\cos x} \right)^n}$ i.e. for $n \geqslant 1$, we can say that ${\cos ^n}x = {\left( {\cos x} \right)^n}$. But this is not the case with negative powers. For $n < 0$, ${\cos ^n}x$ and ${\left( {\cos x} \right)^n}$ are different. And, so we need to check for these cases, then decide whether ${\cos ^n}x = {\left( {\cos x} \right)^n}$ or ${\left( {\cos x} \right)^n} \ne {\cos ^n}x$ for the given $n$.