Prove that, (3-4sin2A)(sec2A-4tan2A)=(3-tan2A)( 1-4sin2A)​

(3-4sin^2x) (1-3tan^2x)

= (3-4sin^2x) * (cos^2x-3sin^2x)/cos^2x

= (3-4sin^2x)/cos^2x * (cos^2x-3sin^2x)

= (3sec^2x-4tan^2x)(cos^2x-3sin^2x)

= (3+3tan^2x-4tan^2x)(cos^2x-3sin^2x)

= (3-tan^2x)(cos^2x-3sin^2x)

= (3-tan^2x)(cos^2x-3+3cos^2x)

= (3-tan^2x)(4cos^2x-3)

Step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length