Question

(1 + tan²A) (1 + cot²A) = 1/sin²A - sin⁴A ​

Answer 1

LHS = ( 1 + 1/tan² A ) + ( 1 + 1/Cot²A )

= ( 1 + Cot²A ) ( 1 + Tan²A )

=> Cosec²A × Sec²A

=> 1/Sin²A × 1/Cos²A

=> 1/Sin²A Cos²A

=> 1/Sin²A ( 1 - Sin²A )

=> 1/ Sin²A - Sin⁴A ) = RHS

Hence,

LHS = RHS = 1/ Sin² A - Sin⁴A .