Question

Prove that :-
(cot³A.sin³A)/(cosA+sinA)² + (tan³A.cos³A)/(cosA+sinA)² = (secA.cosecA-1)/(cosecA+secA)​

Answer 1

Answer:

LHS =

tan³A/1+tan²A + cot³A /++cot²A

= sin³A/cosA +cos³A/sinA ( ∵tanA = sinA/ cosA,)

= sin⁴A+cos⁴A/ cosA sinaA

=sin⁴A+cos⁴A+sin²A×cos²A -sin²A×cos²A

=(sin²A+cos²A) -sin²A×cos²A / sinA×cosA

= 1 -2sin²A ×cos²A / sinA×cosA

= secA·cosecA - 2sinA·cosA (∵ secA = 1/ cosA, cosecA = 1/ sinA)

Hence LHS = RHS