Answer
We want to verify the following identity:
$(1 - sin(x))/(1 + sin(x)) = (sec(x) - tan(x))^2$
Work Step by Step
Verify the following identity:
$(1 - sin(x))/(1 + sin(x)) = (sec(x) - tan(x))^2$
Multiply both sides by $1 + sin(x)$:
$1 - sin(x) = ^?(1 + sin(x)) (sec(x) - tan(x))^2$
Write secant as 1/cosine and tangent as sine/cosine:
$1 - sin(x) = ^?(1 + sin(x)) (1/(cos(x)) - (sin(x))/(cos(x)))^2$
Put $1/(cos(x)) - (sin(x))/(cos(x))$ over the common denominator $cos(x): 1/(cos(x)) - (sin(x))/(cos(x)) = (1 - sin(x))/(cos(x))$:
$1 - sin(x) = ^?(1 - sin(x))/(cos(x))^2 (1 + sin(x))$
Multiply each exponent in $(1 - sin(x))/(cos(x))$ by $2$:
$1 - sin(x) = ^?((1 - sin(x))^2)/(cos(x)^2) (1 + sin(x))$
Multiply both sides by $cos(x)^2$:
$cos(x)^2 (1 - sin(x)) = ^?(1 - sin(x))^2 (1 + sin(x))$
$(1 - sin(x)) cos(x)^2 = cos(x)^2 - cos(x)^2 sin(x)$:
$cos(x)^2 - cos(x)^2 sin(x) = ^?(1 - sin(x))^2 (1 + sin(x))$
Divide both sides by $sin(x) - 1$:
$-cos(x)^2 = ^?(sin(x) - 1) (1 + sin(x))$
$cos(x)^2 = 1 - sin(x)^2$:
$-1 - sin(x)^2 = ^?(sin(x) - 1) (1 + sin(x))$
$-(1 - sin(x)^2) = sin(x)^2 - 1$:
$sin(x)^2 - 1 = ^?(sin(x) - 1) (1 + sin(x))$
$(sin(x) - 1) (1 + sin(x)) = sin(x)^2 - 1$:
$sin(x)^2 - 1 = ^?sin(x)^2 - 1$
The left hand side and right hand side are identical, thus verifying the identity.
