Question

only genious can solve this
4(cos^4A + sin^4A) / (3+cos4A) = (cos^4A-sin^4A) sec2A?

Answer 1

(sin²A+cos²A)²  

as (a+b)²=a²+b²+2ab

(sin²A+cos²A)²=sin^4A+cos^4A+2sin²Acos²A        use sin²a+cos²a=1

(1)² - 2sin²Acos²A = sin^4A+cos^4A

1-2sin²Acos²A = sin^4A+cos^4A

hence proved