Tan³A/1+tan²A + cot³A/1+cot²A=secA cosecA- 2 sinA cos A
Answers
Answered by
3
Solution:
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
Similar questions