Prove that tan^2A-tan^2B=sin(A+B).sin(A-B)/cos^2A.cos^2B

 Solution:-

How to Prove:- 

• To prove tan^2A-tan^2B=sin(A+B).sin(A-B)/cos^2A.cos^2B, we have to take R.H.S instead of L.H.S.

• Next step involves the formula of sin(A+B).sin(A-B)=sin^2A-sin^2B

• After that fraction separation take place as a result it becomes sin^2A/cos^2A.cos^2B-sin^2B/cos^2A.cos^2B

• we know that sinA/cosA=tanA similarly sin^2A/cos^2A=tan^2A and sin^2B/cos^2B=tan^2B and remaining 1/cos^2B and 1/cos^2A becomes sec^2B and sec^2A respectively.

• so the equation becomes tan^2A.sec^2B-tan^2B.sec^2A, put sec^2B=1+tan^2B and sec^2B=1+tan^2B {sec^2x-tan^2x=1  => sec^2x=1+tan^2x}

• Now multiply, find +tan^2A.tan^2B and -tan^2A.tan^2B, they got canceld and we have the required L.H.S that is tan^2A-tan^2B

• Hence L.H.S=R.H.S, Proved.