Question

Trigonometric Ratios

Sketch a triangle that has acute angle $\mathit{\theta }$, and find the other five trigonometric ratios of $\mathit{\theta }$

$\mathit{c}\mathit{o}\mathit{t}\mathit{\theta }\mathbf{=}\mathbf{1}$

Short answer

The required triangle is

The other ratios are-

$\begin{array}{l}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\\ \mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }\mathbf{=}\mathbf{1}\\ \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\\ \mathbf{c}\mathbf{s}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{1}}\\ \mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{1}}\end{array}$

Step-by-step solution

Step 1. Given information

A trigonometric ratio is given as $\cot \theta =1$.

Step 2. Concept used

First find the unknown side of triangle. And then sketch the triangle by using the following trigonometric ratios-

$\begin{array}{l}\Rightarrow \sin \theta =\frac{p}{h}\&\csc \theta =\frac{h}{p}\\ \Rightarrow \cos \theta =\frac{b}{h}\&\sec \theta =\frac{h}{b}\\ \Rightarrow \tan \theta =\frac{p}{b}\&\cot \theta =\frac{b}{p}\end{array}$

Where

$\begin{array}{l}h=hypotenuse\\ b=base\\ p=perpendicular\end{array}$

Step 3. Calculation

The triangle by using trigonometric ratios is-

Perpendicular is given as 1 and base is 1. Now use Pythagoras theorem for finding the third side.

$\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(x\right)}^2={\left(1\right)}^2+{\left(1\right)}^2\\ x^2=1+1\\ x^2=2\\ x=\sqrt{2}\end{array}$

Now we know all three sides. And the other five ratios are-

$\begin{array}{l}\sin \theta =\frac{1}{\sqrt{2}}\\ \tan \theta =1\\ \cos \theta =\frac{1}{\sqrt{2}}\\ \csc \theta =\frac{\sqrt{2}}{1}\\ \sec \theta =\frac{\sqrt{2}}{1}\end{array}$