Proving $\frac{\cot\theta-\csc\theta+1}{\cot\theta+\csc\theta-1}=\csc\theta-\cot\theta$

Question

Prove that $$\frac{\cot\theta - \csc\theta + 1}{\cot\theta+\csc\theta-1}= \csc\theta-\cot\theta$$

Answer 1

$$\operatorname{cosec}^2\theta - \cot^2\theta = 1 \implies (\operatorname{cosec}\theta+\cot\theta)(\operatorname{cosec}\theta-\cot\theta) = 1$$

$$\operatorname{cosec}\theta+\cot\theta -1= \frac{1}{\operatorname{cosec}\theta-\cot\theta}-1 = \frac{1-(\operatorname{cosec}\theta-\cot\theta)}{\operatorname{cosec}\theta-\cot\theta} = \frac{\cot\theta-\operatorname{cosec}\theta+1}{\operatorname{cosec}\theta-\cot\theta} $$

$$\frac{\cot\theta +\operatorname{cosec}\theta-1}{\cot\theta-\operatorname{cosec}\theta+1}=\frac{1}{\operatorname{cosec}\theta-\cot\theta}$$

$$\frac{\cot\theta-\operatorname{cosec}\theta+1}{\cot\theta+\operatorname{cosec}\theta -1} = \operatorname{cosec}\theta-\cot\theta$$

Answer 2

$$\cot\theta-\csc\theta+1$$ $$=\cot\theta-\csc\theta+(\csc^2\theta-\cot^2\theta)$$

$$=(\cot\theta-\csc\theta)(1-\cot\theta-\csc\theta)$$

$$=-(\cot\theta-\csc\theta)(\cot\theta+\csc\theta-1)$$

Answer 3

Another way:

Write $\theta=2y$

$$\cot2y-\csc2y=-\dfrac{1-\cos2y}{\sin2y}=-\tan y$$

$$\cot2y+\csc2y=-\dfrac1{\cot2y-\csc2y}=\cot y$$