Question

Prove that,
tan^2Φ - sin^2Φ = tan^2Φ x sin^2Φ

Answer 1

tan^2A - sin^2A = tan^2 A × sin^2 A

On LHS

tan^2 A - sin^2 A

= sin^2 A / cos^2 A - sin^2 A

= sin^2 A - cos^2 A . sin^2 A / cos^2 A

= sin^2 A ( 1 - cos^2 A ) / cos^2 A

= sin^2 A × sin^2 A / cos^2 A

= sin^4 A / cos^2 A

On RHS

tan^2 A × sin^2 A

= sin^2 A / cos^2 A × sin^2 A

= sin^4 / cos^2 A

so LHS = RHS

SO

tan^2 A - sin^2 A = tan^2 A × sin^2 A

Answer 2

Hello friend

_________________________

Take L.H.S

tan^2¢ - sin^2¢

===> Sin^2¢/cos^2 - sin^2¢

===> sin^2¢ - sin^2¢ x cos^2¢/cos^2¢

===> sin^2¢(1- cos^2¢)/cos^2¢

===> sin^2¢ x sin^2¢/cos^2¢

===> tan^2¢ x sin^2¢

==> L.H.S = R.H.S

We have proved..

Thanks.