Question

Prove that tan^2Φ-sin^2Φ=tan^2Φ*sin^2Φ​

Answer 1

Answer:

$hello$

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

Answer:

(sin/cos)^2-sin^2

sin^2/cos^2-sin^2

sin^2-sin^2*cos^2/cos^2

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

sin^4/cos^2

RHS=

sin^2*sin^2/cos^2

sin^4/cos^2

HP