Establish the identity. ((cot^2 x)/(csc(x)-1)) = (1+sin(x))/(sin(x)) = ____________________ Could someone explain to me how to solve this?

Aug 3, 2018

Explanation:

We have: $\frac{{\cot}^{2} \left(x\right)}{\csc \left(x\right) - 1}$

One of the Pythagorean identities is ${\cot}^{2} \left(x\right) + 1 = {\csc}^{2} \left(x\right)$.

We can rearrange it to get:

$R i g h t a r r o w {\cot}^{2} \left(x\right) = {\csc}^{2} \left(x\right) - 1$

Let's apply this rearranged identity to our proof:

$= \frac{{\csc}^{2} \left(x\right) - 1}{\csc \left(x\right) - 1}$

The numerator is the difference of two squares, and can be factorised as:

$= \frac{\left(\csc \left(x\right) + 1\right) \left(\csc \left(x\right) - 1\right)}{\csc \left(x\right) - 1}$

$= \csc \left(x\right) + 1$

Now, one of the standard trigonometric identities is $\csc \left(x\right) = \frac{1}{\sin \left(x\right)}$.

Applying this, we get:

$= \frac{1}{\sin \left(x\right)} + 1$

$= \frac{1 + \sin \left(x\right)}{\sin \left(x\right)} \text{ " }$ $\text{Q.E.D.}$