The answer is, sec2A + tan2A − 2tan
Answer:
Consider the L.H.S,
=sec4A(1−sin4A)−2tan2A
=sec4A−sec4Asin4A−2tan2A
=sec4A−cos4Asin4A−2tan2A
=sec4A−tan4A−2tan2A
=[(sec2A)2−(tan2A)2]−2tan2A
=[(sec2A)−(tan2A)][(sec2A)+(tan2A)]−2tan2A
We know that
sec2A−tan2A=1
Therefore,
=sec2A+tan2A−2tan
7.9m questions
10.6m answers