Question

Prove the quotient identities given in formula (3).

Short answer

By using the trigonometry ratios in right angled triangle, we proved the following quotient identities :-

(a) $\tan \theta =\frac{\sin \theta }{\cos \theta }$

(b) $cot\theta =\frac{\cos \theta }{\sin \theta }$

Step-by-step solution

Step 1. Given Information

Here we have to prove the following quotient identities :-

(a) $\tan \theta =\frac{\sin \theta }{\cos \theta }$

(b) $cot\theta =\frac{\cos \theta }{\sin \theta }$

We will apply trigonometry ratios, to prove these quotient identities.

Step 2. To prove identity (a) tanθ=sinθcosθ

Consider the following right angled triangle.

For this right angled triangle, sine function is defined as perpendicular upon hypotenuse.

That is :-

$\sin \theta =\frac{AB}{AC}$$..........\left(1\right)$

Also cosine function is defined as base upon hypotenuse.

That is :-

localid="1647181073885" $\cos \theta =\frac{BC}{AC}$$.........\left(2\right)$

Now tangent function is defined as perpendicular upon base.

That is :-

$\tan \theta =\frac{AB}{BC}$$.........\left(3\right)$

Divide $\left(1\right)by\left(2\right)$, then we have :-

role="math" localid="1647181511822" $$\begin{gathered} \frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \frac{AB}{AC}}}{{\displaystyle \frac{BC}{AC}}} \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{AB}{AC}\times \frac{AC}{BC} \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{AB}{BC}............\left(3\right) \end{gathered}$$

By comparing $\left(2\right)and\left(3\right)$, we have :-

The right hand sides of both equations are equal, so left hand sides are also equal. This gives us :-

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

Hence proved.

Step 3. To prove identity (b) cotθ=cosθsinθ

Consider the following right angled triangle.

For this right angled triangle, cotangent function is defined as base upon perpendicular.

That is :-

$cot\theta =\frac{BC}{AB}$$.........\left(4\right)$

Divide $\left(2\right)by\left(1\right)$, then we have :-

localid="1647182332148" $$\begin{gathered} \frac{\cos \theta }{\sin \theta }=\frac{{\displaystyle \frac{BC}{AC}}}{{\displaystyle \frac{AB}{AC}}} \\ \Rightarrow \frac{\cos \theta }{\sin \theta }=\frac{BC}{AC}\times \frac{AC}{AB} \\ \Rightarrow \frac{\cos \theta }{\sin \theta }=\frac{BC}{AB}...........\left(5\right) \end{gathered}$$

Now by comparing $\left(4\right)and\left(5\right)$, we have :-

The right hand sides of both equations are equal, so left hand sides are also equal. This gives us :-

$cot\theta =\frac{\cos \theta }{\sin \theta }$

Hence proved.