# How do you perform multiplication and use the fundamental identities to simplify (cotx+cscx)(cotx-cscx)?

Aug 22, 2017

$- 1$

#### Explanation:

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

Applying that here we have

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

now the identity connecting $\cot x \text{ & "cscx" is as follows}$

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

substituting for ${\csc}^{2} x$

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

$= \cancel{{\cot}^{2} x - {\cot}^{2} x} - 1$

$= - 1$

Aug 22, 2017

$- 1$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)cotx=cosx/sinx" and "cscx=1/sinx

•color(white)(x)cos^2x=1-sin^2x

$\text{expand the factors using the FOIL method}$

$\Rightarrow \left(\cot x + \csc x\right) \left(\cot x - \csc x\right)$

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

$= {\cos}^{2} \frac{x}{\sin} ^ 2 x - \frac{1}{\sin} ^ 2 x \leftarrow \text{ common denominator}$

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

$= \frac{\cancel{1} - {\sin}^{2} x \cancel{- 1}}{\sin} ^ 2 x$

$= - {\cancel{{\sin}^{2} x}}^{1} / {\cancel{{\sin}^{2} x}}^{1}$

$= - 1$