Question

Find out the values of angle $-135°$ for all the six trigonometric ratios.

Answer 1

Step 1: Compute the $\text{sine}$ and $\mathrm{cosine}$ of the given angle.

In the question, the measure of an angle $-135°$ is given.

Compute the $\text{sine}$ of the given angle.

We know that, $\sin (-\theta )=-\sin \left(\theta \right)$ and $\sin \left(180°-\theta \right)=\sin \left(\theta \right)$.

So,

$$\begin{gathered} \sin (-135°)=-\sin (135°) \\ \Rightarrow \sin (-135°)=-\sin (180°-45°) \\ \Rightarrow \sin (-135°)=-\sin (45°) \\ \Rightarrow \sin (-135°)=-\frac{1}{\sqrt{2}} \end{gathered}$$

Compute the $\text{cosine}$ of the given angle.

We know that, $\cos (-\theta )=\cos \left(\theta \right)$ and $\cos \left(180°-\theta \right)=-\cos \left(\theta \right)$.

So,

$$\begin{gathered} \cos (-135°)=\cos (135°) \\ \Rightarrow \cos (-135°)=\cos (180°-45°) \\ \Rightarrow \cos (-135°)=-\cos (45°) \\ \Rightarrow \cos (-135°)=-\frac{1}{\sqrt{2}} \end{gathered}$$

Step 2: Compute the $\text{tangent}$ and $\mathrm{cotangent}$ of the given angle.

Compute the $\text{tangent}$ of the given angle.

We know that, $\tan (-\theta )=-\tan \left(\theta \right)$ and $\tan \left(180°-\theta \right)=-\tan \left(\theta \right)$.

So,

$$\begin{gathered} \tan (-135°)=-\tan (135°) \\ \Rightarrow \tan (-135°)=-\tan (180°-45°) \\ \Rightarrow \tan (-135°)=\tan (45°) \\ \Rightarrow \tan (-135°)=1 \end{gathered}$$

Compute the $\text{cotangent}$ of the given angle.

We know that, $cot(-\theta )=-cot\left(\theta \right)$ and $cot\left(180°-\theta \right)=-cot\left(\theta \right)$.

So,

$$\begin{gathered} cot(-135°)=-cot(135°) \\ \Rightarrow cot(-135°)=-cot(180°-45°) \\ \Rightarrow cot(-135°)=cot(45°) \\ \Rightarrow cot(-135°)=1 \end{gathered}$$

Step 3: Compute the $\text{secant}$ and $\text{cosecant}$ of the given angle.

Compute the $\text{secant}$ of the given angle.

We know that, $sec(-\theta )=sec\left(\theta \right)$ and $sec\left(180°-\theta \right)=-sec\left(\theta \right)$.

So,

$$\begin{gathered} sec(-135°)=sec(135°) \\ \Rightarrow sec(-135°)=sec(180°-45°) \\ \Rightarrow sec(-135°)=-sec(45°) \\ \Rightarrow sec(-135°)=-\sqrt{2} \end{gathered}$$

Compute the $\text{cosecant}$ of the given angle.

We know that, $cosec(-\theta )=-cosec\left(\theta \right)$ and $\cos ec\left(180°-\theta \right)=\cos ec\left(\theta \right)$.

So,

$$\begin{gathered} \cos ec(-135°)=-cosec(135°) \\ \Rightarrow \cos ec(-135°)=-cosec(180°-45°) \\ \Rightarrow \cos ec(-135°)=-cosec(45°) \\ \Rightarrow \cos ec(-135°)=-\sqrt{2} \end{gathered}$$

Hence, the values of trigonometric ratios are

$$\begin{gathered} \sin (-{135}^{\circ })=-\frac{1}{\sqrt{2}} \\ \cos (-{135}^{\circ })=-\frac{1}{\sqrt{2}} \\ \tan (-{135}^{\circ })=1 \\ \cos ec(-{135}^{\circ })=-\sqrt{2} \\ sec(-{135}^{\circ })=-\sqrt{2} \\ cot(-{135}^{\circ })=1 \end{gathered}$$