Question

Solve the problem quickly with full process

Answer 1

from L.H.S.

=2sin{2A+2B}/2×COS{2A-2B}/2/2COS{2A+2B}/2×SIN{2A-2B}/2

=2sin{A+B}×COS{A-B}/2COS{A+B}×SIN{A-B}

=SIN{A+B}/COS{A+B}×COS{A-B}/SIN{A-B}

=TAN{A+B}×1/TAN{A-B}

=TAN{A+B}/TAN{A-B}

PROVED R.H.S.

I HOPE THAT THIS ANS WILL BE RYT