Question

Please prove the expression​

Answer 1

Solution:-

To Prove:-

[ (sin²A - sin²B)/(sinA.cosA - sinB.cosB)] = tan ( A + B)

Proof :-

=) [(sin²A - sin²B)/(sinA.cosA - sinB.cosB)]

=) [ (1-cos2A)/2 - (1-cos2B)/2 ]/ [ ( 2sinA.cosA)/2 - ( 2sinB.cosB)/2 ]

=) [ (1 - cos2A - 1 + cos2B)/(sin2A - sin2B)]

=) [ ( - cos2A + cos2B)/( sin2A - sin2B) ]

=) [ -(cos2A -cos2B)/(sin2A - sin2B)]

=) { - [ -2sin( 2A + 2B)/2 .sin(2A -2B)/2 ]/[ 2sin( 2A-2B)/2 . cos(2A + 2B)/2 ]}

=) [ sin( A + B)/cos(A+B) ]

=) tan(A + B)

L.H.S. = R.H.S.

Hence Proved!