Question

tan^2a-tan^2b=cos^2b-cos^2a/cos^2bcos^2a=sin^2a-sin^2b/cos^2acos^2b


HELP ME PLEASE

Answer 1

Hope you understand that.

Answer 2

$LHS = tan^2a- tan^2b$

We know that sec^2a = 1 + tan^2a, that means tan^2a = sec^2a - 1.

sec^2a - 1 - (sec^2b - 1)

sec^2a - 1 - sec^2b + 1

sec^2a - sec^2b

1/cos^2a - 1/cos^2b

On cross- multiplication, we get

(cos^2b - cos^2a)/(cos^2b * cos^2a)

(1 - sin^2b - (1 - sin^2a)/(cos^2b * cos^2a)

(1 - sin^2b - 1 + sin^2a)/(cos^2b * cos^2a)

(sin^2a - sin^2b)/(cos^2b * cos^2a)


LHS = RHS.


Hope this helps!