Question

prove that (tan³A/1+tan²A) + (cot³A/1+cot²A) = secAcosecA - 2sinAcosA​

Answer 1

Answer:

=Tan³A/1+tan²A + cot³A/1+cot²A

=secA cosecA- 2 sinA cos A

=tan³A/(1+tan²A) + cot²A/(1+cot²A)

=tan³A/sec²A + cot³A/csc²A

=sin³A/cosA + cos³A/sinA

=(sin⁴A + cos⁴A)/(sinA.cosA)

=[(sin²A + cos²A)² -2(sin²A.cos²A)/(sinA.cosA)

=(1 - 2sin²A.cos²A)/(sinA.cosA)

Step-by-step explanation:

Answer 2

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