Question

Simplify this equation please.

Answer 1

$\dfrac{\csc^2\theta-3\csc\theta+2}{\csc^2\theta-1}$

Identity:

$\sin^2\theta+\cos^2\theta=1\implies1+\cot^2\theta=\csc^2\theta$

So we can rewrite the denominator to get

$\dfrac{\csc^2\theta-3\csc\theta+2}{\cot^2\theta}$

Multiply numerator and denominator by $\sin^2\theta$. Several terms will cancel since $\sin\theta\csc\theta=1$. Also, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$. We get

$\dfrac{1-3\sin\theta+2\sin^2\theta}{\cos^2\theta}$

Factorize the numerator, and write $\cos$ in terms of $\sin$ in the denominator to factorize it further to get

$\dfrac{(1-\sin\theta)(1-2\sin\theta)}{\cos^2\theta}=\dfrac{(1-\sin\theta)(1-2\sin\theta)}{1-\sin^2\theta}=\dfrac{(1-\sin\theta)(1-2\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}$


The $1-\sin\theta$ factors cancel, leaving you with

$\dfrac{1-2\sin\theta}{1+\sin\theta}$

which you could simplify a bit further by writing

$\dfrac{1+\sin\theta-3\sin\theta}{1+\sin\theta}=1-\dfrac{3\sin\theta}{1+\sin\theta}$