Question

simplique as expressoes:a) sec x - cossec x /1-cotg xb)y=(sec x - cos x) (cossec x - sen x) (tg x + cotg x)como fazer

Answer 1

a)$\frac{(secx-cossecx)}{1-cotgx}$
Sabemos que:
$cotgx = tg^{-1}=\frac{cosx}{senx}$
$secx=\frac{1}{cosx}$
$cossecx=\frac{1}{senx}$
Basta substituirmos e gozarmos a álgebra heueheuheueh:
$\frac{\frac{1}{cosx}-\frac{1}{senx}}{1-\frac{cosx}{senx}}$
$\frac{\frac{(senx-cosx)}{senxcosx}}{\frac{(senx-cosx)}{senx}}$
$\frac{senx}{senxcosx}$
$\frac{1}{cosx}=secx$