Verify the trigonometric identity square root [(1+sin x)/(1-sinx)]=(1+sinx)/|cosx|

We'll manage the left side of the expression and we'll
multiply the numerator and denominator by (1+sin x).

sqrt
[(1+sin x)^2/(1-sin x)*(1+sin x)] = (1+sinx)/|cosx|

We'll
recognize the difference of 2 squares at denominator:

sqrt
[(1+sin x)^2/[1- (sin x)^2]

We'll apply the Pythagorean
identity:

1- (sin x)^2 = (cos
x)^2

sqrt [(1+sin x)^2/[1- (sin x)^2] = |(1+sin x)^2|/|cos
x|

Since 1 + sin x >= 0 => |(1+sin x)| =
(1+sin x)

sqrt [(1+sin x)^2/[1- (sin x)^2] = (1+sin x)/|cos
x|

We notice that LHS = RHS, therefore the
identity sqrt [(1+sin x)^2/[1- (sin x)^2] = (1+sin x)/|cos x| is
verified.