# Question #8de97

Dec 10, 2016

That is not an identity.

#### Explanation:

Recall that

${\cot}^{2} x + 1 = {\csc}^{2} x$.

So, we can write

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

$= {\cot}^{2} \frac{x}{\csc} ^ 2 x$

Recall also that $\cot x = \cos \frac{x}{\sin} x$ and $\csc x = \frac{1}{\sin} x$.

This allows us to continue

$= \frac{{\cos}^{2} \frac{x}{\sin} ^ 2 x}{\frac{1}{\sin} ^ 2 x}$

$= {\cos}^{2} \frac{x}{\sin} ^ 2 x \cdot {\sin}^{2} \frac{x}{1}$

$= {\cos}^{2} x$

Which is not identically $\cos x$.

(${\cos}^{2} x = \cos x$ only when $\cos x = 1$ or $0$)

Dec 10, 2016

No. It is equal to ${\sin}^{2} x - 1$.

#### Explanation:

If we have $\frac{1 - x}{x}$, we can write it as $\frac{1}{x} - \frac{x}{x}$.

The same way, $\frac{1 - {\csc}^{2} x}{\csc} ^ 2$ can be written as $\frac{1}{\csc} ^ 2 x - \frac{{\csc}^{2} x}{{\csc}^{2} x}$.

This is equal to ${\sin}^{2} x - 1$.