# Simplify ?  (cos^2a - sin^2b)/(sin^2a • sin^2b) - ctg^2a • ctg^2b

Apr 5, 2018

$\frac{{\cos}^{2} \left(a\right) - {\sin}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)} - {\cot}^{2} \left(a\right) {\cot}^{2} \left(b\right) = - 1$

#### Explanation:

We want to simplify

$\frac{{\cos}^{2} \left(a\right) - {\sin}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)} - {\cot}^{2} \left(a\right) {\cot}^{2} \left(b\right)$

We will use the identity

• ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

Thus

$\frac{{\cos}^{2} \left(a\right) - {\sin}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)} - {\cot}^{2} \left(a\right) {\cot}^{2} \left(b\right)$

$\frac{{\cos}^{2} \left(a\right) - {\sin}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)} - \frac{{\cos}^{2} \left(a\right) {\cos}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)}$

$\frac{{\cos}^{2} \left(a\right) - {\sin}^{2} \left(b\right) - {\cos}^{2} \left(a\right) {\cos}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)}$

$\frac{{\cos}^{2} \left(a\right) \left(1 - {\cos}^{2} \left(b\right)\right) - {\sin}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)}$

$\frac{{\cos}^{2} \left(a\right) {\sin}^{2} \left(b\right) - {\sin}^{2} \left(b\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)}$

$\frac{- {\sin}^{2} \left(b\right) \left(1 - {\cos}^{2} \left(a\right)\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)}$

$\frac{- {\sin}^{2} \left(b\right) {\sin}^{2} \left(a\right)}{{\sin}^{2} \left(a\right) {\sin}^{2} \left(b\right)}$

$- 1$