Question

1-tan²a/1+tan²a = 2cos²a-1

Answer 1

LHS: (1+tan²A)/(1+cot²A) =(1+tan²A)/(1+1/tan²A) =( 1+tan²A)/[(1+tan²A)/tan²A] =tan²A. RHS: (1-tanA)²/(1 -cotA)² =(1-2tanA+tan²A)/(1-2cotA+cot²A)