Question

Express cot^2 theta (4cosec^2 theta + tan^2 theta) as perfect square​

Answer 1

Cot^2@(4Cosec^2@+tan^2@)
Cot^2@(4Cosec^2@+Sin^2@/Cos^2@)
Cot^2@(4Cosec^2@+Cot^2@/Cosec^2@)
Cot^2@(4+Cot^2@)
4Cot^2@+Cot^4@

Or,
(2Cot@)^2+(Cot^2@)^2

Hope it will help you!!

Answer 2

Answer:

Let theta be A.

(cosec²A + cot²A)²

Step-by-step explanation:

cot²(4cosec²A + tan²A)

=cos²A/sin²A(1/sin²A + sin²A/cos²A)

=4cos²A/sin⁴A + 1

=(4 - 4sin²A )/sin⁴A + 1

=4/sin⁴A - 4sin²A/sin⁴A + 1

=4cosec⁴A - 4/sin²A + 1

=4cosec⁴A - 4cosec²A + 1

=( 2cosec²A - 1 )²

=( cosec²A + cosec²A + 1 )²

=( cosec²A + cot²A )²