# How do I prove and find the domain of the following trig identity?

## $\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = \frac{\cot \theta - 1}{\cot \theta + 1}$

Nov 22, 2016

see below

#### Explanation:

$\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = \frac{\cot \theta - 1}{\cot \theta + 1}$

Right Side : $= \frac{\cot \theta - 1}{\cot \theta + 1}$

$= \frac{\cos \frac{\theta}{\sin} \theta - 1}{\cos \frac{\theta}{\sin} \theta + 1}$

=((cos theta-sin theta)/sin theta)/((cos theta+sin theta)/sin theta

$= \frac{\cos \theta - \sin \theta}{\sin} \theta \cdot \sin \frac{\theta}{\cos \theta + \sin \theta}$

$= \frac{\cos \theta - \sin \theta}{\cancel{\sin}} \theta \cdot \cancel{\sin} \frac{\theta}{\cos \theta + \sin \theta}$

$= \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$

$\therefore =$ Left Side

Domain :
The denominator cannot be zero. So take the denominator set it equal to zero and then solve.

$\cos \theta + \sin \theta = 0$

Use linear combination to solve the equation

$A = 1 , B = 1 , C = \sqrt{2}$. Note that this is in quadrant one since both cosine and sine are positive

$\cos D = \frac{A}{C} = \frac{1}{\sqrt{2}}$

$D = {\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = {45}^{\circ}$

$\sqrt{2} \cos \left(\theta - {45}^{\circ}\right) = 0$--> Put it in the form $C \cos \left(\theta - D\right) = 0$

$\cos \left(\theta - {45}^{\circ}\right) = 0$

$\theta - {45}^{\circ} = {\cos}^{-} 1 0$

$\theta - {45}^{\circ} = \pm {90}^{\circ} + {360}^{\circ} n$

$\theta = {45}^{\circ} \pm {90}^{\circ} + {360}^{\circ} n$

$\theta = {135}^{\circ} + {360}^{\circ} n \mathmr{and} \theta = - {45}^{\circ} + {360}^{\circ} n$

where $n = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

Therefore,

$D : \left\{\theta \in R , \theta \ne {135}^{\circ} + {360}^{\circ} n , \theta \ne - {45}^{\circ} + {360}^{\circ} n , n = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots\right\}$