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)?

2 Answers
Dec 7, 2017

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)

Dec 7, 2017

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.