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

### 3 Answers | Add Yours

We have to prove that (tan x)^2 - (sin x)^2 = (tan x)^2 * (sin x)^2

Start from the left hand side:

(tan x)^2 - (sin x)^2

use tan x = sin x / cos x

(sin x)^2/(cos x)^2 - (sin x)^2

=> [(sin x)^2 - (sin x)^2*(cos x)^2]/(cos x)^2

=> (sin x)^2[1 - (cos x)^2]/(cos x)^2

=> (sin x)^2 * [(sin x)^2 / (cos x)^2]

=> (tan x)^2 * (sin x)^2

which is the right hand side.

**This proves that (tan x)^2 - (sin x)^2 = (tan x)^2 * (sin x)^2**

R:H:S ≡ tan²x.sin²x

= tan²x(1-cos²x)

= tan²x - tan²x.cos²x

= tan²x - (sin²x/cos²x)*cos²x

= tan²x - sin²x

= L:H:S

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.**