Proving Trig Identity?

L.H.S.
= cot²A + csc²A
= (cosA/sinA)² + (1/sinA)²
= (cos²A/sin²A) + (1/sinA²)
= (cos²A + 1)/sin²A
= [(1 + cos²A)/sin²A] × (sin²A/sin²A)
= [(1 + cos²A)/sin²A] × [(1 - cos²A)/sin²A]
= (1 + cos²A)(1 - cos²A)/sin⁴A
= (1 - cos⁴A)/sin⁴A
= (1/sin⁴A) - (cos⁴A/sin⁴A)
= -(cosA/sinA)⁴ + (1/sinA)⁴
= -cot⁴A + csc⁴A
= R.H.S.

Hence, cot²A + csc²A = -cot⁴A + csc⁴A

= - cot⁴(a) + csc⁴(a)

= csc⁴(a) - cot⁴(a)

= [csc²(a)]² - [cot²(a)]² → recall: u² - v² = (u + v).(u - v)

= [csc²(a) + cot²(a)].[csc²(a) - cot²(a)] → recall: csc(a) = 1/sin(a)

= [{1/sin(a)}² + cot²(a)].[{1/sin(a)}² - cot²(a)] → recall: cot(a) = cos(a)/sin(a)

= [{1/sin(a)}² + {cos(a)/sin(a)}²].[{1/sin(a)}² - {cos(a)/sin(a)}²]

= [{1/sin²(a)} + {cos²(a)/sin²(a)}].[{1/sin²(a)} - {cos²(a)/sin²(a)}]

= [{1 + cos²(a)}/sin²(a)].[{1 - cos²(a)}/sin²(a)] → recall: 1 - cos²(a) = sin²(a)

= [{1 + cos²(a)}/sin²(a)].[sin²(a)/sin²(a)] → you can simplify

= [1 + cos²(a)]/sin²(a)

= [1/sin²(a)] + [cos²(a)/sin²(a)]

= [1/sin(a)]² + [cos(a)/sin(a)]² → recall: 1/sin(a) = csc(a)

= csc²(a) + [cos(a)/sin(a)]² → recall: cos(a)/sin(a) = cot(a)

= csc²(a) + cot²(a)

cot^2 A + csc^2 A = - cot^4 A + csc^4 A

LHS  =  cot^2 A + csc^2 A  =  cot^2 A +  ( 1 + cot^2 A ) = ( 1 + 2 cot^2 A )

Right Hand Side  =  - cot^4 A + csc^4 A  =  - cot⁴A + (1+cot^2A)^2

=> RHS  =  - cot⁴A + ( 1 + cot^4A+ 2 cot^2A) = 1 + cot^4A  - cot^4A + 2 cot^2A

=>RHS = ( 1 + 2 cot^2 A )

Hence  LHS  =  RHS  Proved 

Remember the formula :  csc^2 A  = (1 + cot^2A )