Consider the L.H.S,
=sec4A(1−sin4A)−2tan2A
=sec4A−sec4Asin4A−2tan2A
=sec4A−sin4Acos4A−2tan2A
=sec4A−tan4A−2tan2A
=[(sec2A)2−(tan2A)2]−2tan2A
=[(sec2A)−(tan2A)][(sec2A)+(tan2A)]−2tan2A
We know that
sec2A−tan2A=1
Therefore,
=sec2A+tan2A−2tan2A
=sec2A−tan2A
=1
Hence, proved