# Verify: tan^2x - sin^2x= (tan^2x)(sin^2x)

### 3 Answers | Add Yours

tan^2x - sin^2x =

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

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

We know the `sin^2+cos^x = 1` .

Therefore `sin^2x = 1-cos^2x` .

`Tanx = sinx/cosx.` We use these identities in the course of the solution.

RHS: `tan^2 x * sin^2x = tan^2x (1-cos^2x)` ,

`=tan^2x-(tan^2x)*cosx^2` .

`=tan^2x-(sinx/cosx)^2*cos^2x.`

`=tan^2 x - (sin^2x/cos^2x)cos^2x`

`=tan^2x - sin^2x*cos^2x/cos^2x`

`=tan^2x - sin^2 x ` which is LHS.

R:H:S ≡ tan²x.sin²x

= tan²x(1-cos²x)

= tan²x - tan²x.cos²x

= tan²x - (sin²x/cos²x)*cos²x

= tan²x - sin²x

=R:H:S