Question

Show that: cosec^2 θ - tan^2 θ (90^o - θ) = sin^2 θ + sin^2 (90^o - θ)

Answer 1

Answer:

Step-by-step explanation:

Cosec²∅ - tan²(90-∅) = LHS  

1/sin²∅ - cot²∅

1/sin²∅ - cos²∅/sin²∅

1-cos²∅/sin²∅

sin²∅ / sin²∅

1  

RHS = sin²∅ + sin²(90-∅)  

sin²∅ + cos²∅

1  

hence LHS = RHS