Question

How do you simplify $\sec \left( -x \right)$ ?

Answer 1

To simplify $\sec \left( -x \right)$, first of all, we are going to use the following trigonometric conversion $\sec \theta =\dfrac{1}{\cos \theta }$ . After that we will use the property that $\cos \left( -\theta \right)=\cos \theta$. Combining these two properties of trigonometric functions we can simplify the above trigonometric expression.

Complete answer:
The trigonometric function which we are going to simplify is as follows:
$\sec \left( -x \right)$
Now, we are going to eliminate this secant function by using a trigonometric property of $\sec \theta$ which says that secant of an angle theta is the reciprocal of cosine of that angle theta. In the below, we have written mathematically what we have just stated:
$\sec \theta =\dfrac{1}{\cos \theta }$
So, while applying the above trigonometric secant to cosine conversion in $\sec \left( -x \right)$, we are going to replace $\theta$ by $-x$ we get,
$\Rightarrow \dfrac{1}{\cos \left( -x \right)}$
Now, to further simplify the above expression we need to remove this negative sign which can be eliminated by using the property that the cosine of a negative angle will give the same result as the cosine of the same angle but without a negative sign. The mathematical expression of what we have just stated is shown below:
$\cos \left( -\theta \right)=\cos \theta$
Using the above cosine relation in $\dfrac{1}{\cos \left( -x \right)}$, we are going to replace $\theta$ by $-x$ in the above equation and we get,
$\Rightarrow \dfrac{1}{\cos \left( -x \right)}=\dfrac{1}{\cos x}$
Hence, we have simplified the given expression to $\dfrac{1}{\cos x}$.

Note: In the above solution, we can reduce $\dfrac{1}{\cos x}$ to further by using the property that $\dfrac{1}{\cos x}=\sec x$ so we can write $\dfrac{1}{\cos x}$ as $\sec x$.
From the above solution, we have learnt a concept just like $\cos \left( -x \right)=\cos x$ similarly, $\sec \left( -x \right)=\sec x$ so you can remember this relation which will be helpful in solving larger trigonometric expressions.