Question

Prove the identity tan^2x-sin^2x is same as (tan^2x)(sin^2x)?

Prove the identity `tan^2x-sin^2x`is same as
`(tan^2x)(sin^2x)?`

Answer 1

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

Explanation:

Assuming `tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)`, start off by rewriting `tan^2(x)` in to its `sin(x)` and `cos(x)` components.

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

Next find a common denominator (LCD: `cos^2(x)*1`)

`sin^2(x)/cos^2(x)*(1/1) - sin^2(x)*cos^2(x)/cos^2(x) rarr sin^2(x)/cos^2(x) - (sin^2(x)cos^2(x))/cos^2(x)`

Combine in to a single fraction and factor out a `sin^2(x)`.

`(sin^2(x)-sin^2(x)cos^2(x))/cos^2(x) rarr sin^2(x)*sin^2(x)/cos^2(x)`

Finally just rewrite

`sin^2(x)*sin^2(x)/cos^2(x) rarr sin^2(x)tan^2(x)`

Answer 2

Please refer to a Proof given in the Explanation.

Explanation:

We have,

`tan^2x-sin^2x,`

`=sin^2x/cos^2x-sin^2x,`

`=sin^2x(1/cos^2x-1),`

`=sin^2x{(1-cos^2x)/cos^2x},`

`=sin^2x{sin^2x/cos^2x},`

`=sin^2x*tan^2x.`

Hence, the Proof.