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 \[n\].