Question

((1/cosx)(sinx))/((sinx/cosx-1))

Answer 1

(sinx - cosx + 1)/(sinx + cosx - 1)=(sinx + 1)/cosx
LHS = [sinx - (cosx - 1)]/[sinx + (cosx - 1)]
= [sinx - (cosx - 1)]/[sinx + (cosx - 1)] * [sinx - (cosx - 1)]/[sinx - (cosx - 1)]
= [sinx - (cosx - 1)]^2/[sin^2x - (cosx - 1)^2]
= [sin^2x - 2sinx(cosx - 1) + (cosx - 1)^2]/[sin^2x - (cos^2x - 2cosx + 1)]
= (sin^2x - 2sinx cosx + 2sinx + cos^2x - 2cosx + 1)/(sin^2x - cos^2x + 2cosx - 1)
= (2 - 2sinx cosx + 2sinx - 2cosx)/(1 - 2cos^2x + 2cosx - 1)
= [2(sinx + 1)- 2cosx(sinx + 1)]/[-2cosx(cosx - 1)]
= [-2(sinx + 1) (cosx - 1)]/[-2cosx(cosx - 1)]
= (sinx + 1)/cosx = RHS

Answer 2

well its

\[\frac{\frac{1}{\cos(x)\sin(x)}}{\frac{\sin(x)}{\cos(x) - 1}}\]

the rule for dividing by a fraction is flip the denominator and multiply

\[\frac{1}{\cos(x)\sin(x)} \times \frac{\cos(x) -1}{\sin(x)}\]

Answer 3

so you get

\[\frac{\cos(x) -1}{\cos(x) \sin^2(x)}\]

using sin^2(x) = -(cos^2(x) - 1) and substituting you get

\[\frac{\cos(x) -1}{\cos(x)\times -(\cos^2(x) -1)}\]

which can be written as

\[\frac{\cos(x) -1}{-\cos(x)(\cos^2(x) -1)}\]

factoring the denominator and you get

\[\frac{\cos(x) -1}{-\cos(x)(\cos(x) + 1)(\cos(x) -1)}\]

cancel the common factor and you get

\[\frac{1}{-\cos^2(x) - \cos(x)}\]