Question

What is the exact value of $\sec ({{135}^{\circ }})$ in simplest rationalized form?

Answer 1

Hint: To find the exact value of $\sec ({{135}^{\circ }})$, we will first find the exact value of $\cos ({{135}^{\circ }})$ as trigonometric identities in ‘sine’ and ‘cos’ are much easier to calculate. Once we do that, we can find the inverse of $\cos ({{135}^{\circ }})$. This will give us the required solution and then we could rationalize it to the simplest rational form.

Complete step-by-step solution:
We will first proceed by finding the value of $\cos ({{135}^{\circ }})$ in our problem.
$\cos ({{135}^{\circ }})$ could written as follows:
$\Rightarrow \cos ({{135}^{\circ }})=\cos ({{90}^{\circ }}+{{45}^{\circ }})$
Let us name the above equation as $(1)$, so we have:
$\Rightarrow \cos ({{135}^{\circ }})=\cos ({{90}^{\circ }}+{{45}^{\circ }})$ …….$(1)$
Now, using the trigonometric identity which relates $\cos (x)$ and $\sin (x)$ as:
$\Rightarrow \cos ({{90}^{\circ }}+x)=-\sin (x)$
We can simplify our above equation number $(1)$ as follows:
$\Rightarrow \cos ({{90}^{\circ }}+{{45}^{\circ }})=-\sin ({{45}^{\circ }})$
$\therefore \cos ({{135}^{\circ }})=-\sin ({{45}^{\circ }})$
Let us name the above equation as $(2)$, so we have:
$\Rightarrow \cos ({{135}^{\circ }})=-\sin ({{45}^{\circ }})$ …….$(2)$
Now, we know the value of $\sin ({{45}^{\circ }})$ is equal to $\dfrac{1}{\sqrt{2}}$ . Thus, putting it in equation number $(2)$, we get:
$\Rightarrow \cos ({{135}^{\circ }})=-\left( \dfrac{1}{\sqrt{2}} \right)$
$\therefore \cos ({{135}^{0}})=-\dfrac{1}{\sqrt{2}}$
Let us name the above equation as $(3)$, so we have:
$\Rightarrow \cos ({{135}^{0}})=-\dfrac{1}{\sqrt{2}}$ …….$(3)$
Now, that we have the value of $\cos ({{135}^{0}})$, we can calculate the value of $\sec ({{135}^{0}})$ as follows:
$\Rightarrow \sec ({{135}^{\circ }})=\dfrac{1}{\cos ({{135}^{\circ }})}$
Using the value of $\cos ({{135}^{\circ }})$ from equation number $(3)$, we get:
$\begin{align}
  & \Rightarrow \sec ({{135}^{\circ }})=\dfrac{1}{-\dfrac{1}{\sqrt{2}}} \\
 & \therefore \sec ({{135}^{\circ }})=-\sqrt{2} \\
\end{align}$
Thus, the exact value of $\sec ({{135}^{\circ }})$ comes out to be $-\sqrt{2}$.
As we can see this is already rationalized to its simplest form, thus we do not need to rationalize it any further.
Hence, the exact value of $\sec ({{135}^{0}})$ in simplest rationalized form comes out to be $-\sqrt{2}$ .

Note: It should be noted that problems involving trigonometric identities are comparatively easier and less clumsy to calculate when converted into their ‘sin’ and ‘cos’ counterparts as they are easy to deal with and have simple angle expansion and angle reduction formulas.