Q 130 - Precalculus Enhanced with Graphing Utilities

Question

Prove the quotient identities given in formula (3). 

Step-by-Step Solution

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 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) tan θ = sin θ cos θ

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 :-

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

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 :-

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

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) c o t θ = cos θ sin θ

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 :-

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

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.