Prove: tan^2x - sin^2x = tan^2x sin^2x

We'll recognize to the left side a difference of 2 squares
and we'll re-write it as it follows:

(tan x - sin x)(tan x
+ sin x) = [(tan x)(sin x)]^2

But tan x = sin x/cos
x

sin x/cos x - sin x = sin x(1/cos x - 1) = sin x(1 - cos
x)/cos x (1)

sin x/cos x + sin x = sin x(1/cos x + 1) = sin
x(1 + cos x)/cos x (2)

We'll multiply (1) by
(2):

(sin x)^2*(1 - cos x)(1 + cos x)/(cos x)^2 = (tan
x)^2*(1 - cos x)(1 + cos x)

(sin x)^2*(1 - cos x)(1 + cos
x)/(cos x)^2 = (tan x)^2*[1 - (cos x)^2]

But from
Pythagorean identity, we'll get:

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

LHS = (sin x)^2*(1 - cos x)(1 + cos x)/(cos x)^2 =
(tan x)^2*(sin x)^2 = RHS

Since we've get LHS
= RHS, we can state that the given expression represents an
identity.