Question

Please Answer Me Fast??

Answer 1

got the answer.........

Answer 2

$Given : tan^2A-tan^2B$

$=\frac{sin^2A}{cos^2A}-\frac{sin^2B}{cos^2B}$

$=\frac{sin^2A*cos^2B-cos^2A*sin^2B}{cos^2A*cos^2B}$

$=\frac{(1-cos^2A)*cos^2B-cos^2A*(1-cos^2B)}{cos^2A*cos^2B}$

$=\frac{cos^2B-cos^2Acos^2B-cos^2A+cos^2Acos^2B}{cos^2Acos^2B}$

$=\frac{cos^2B-cos^2A}{cos^2Acos^2B}$

$=\frac{(1-sin^2B)-(1-sin^2A)}{cos^2Acos^2B}$

$=\frac{1-sin^2B-1+sin^2A}{cos^2Acos^2B}$

$=\boxed{\frac{sin^2A-sin^2B}{cos^2Acos^2B}}$

Hope it helps!