Question

(cot A-1)/(2-sec² A)=cot A/(1+tanA)​

Answer 1

Answer:

We know,

⟹ sin²A + cos²A = 1 => tan²A + 1 = sec²A

1 - sec²A = -tan²A.

Hence Denominator of LHS:

⟹ 2 - sec²A = 1 + 1 - sec²A = 1 - tan²A

Numerator of LHS: cotA - 1 = (1/tanA) -1 = (1-tanA)/tanA

Hence LHS:

⟹ (cotA-1) / (2 - sec²A) = [ (1-tanA)/tanA] / 1 - tan²A

=(1-tanA) / [tanA (1+tanA)(1 - tanA)] = 1 / [tanA(1+tanA)]

= (1/tanA) * 1/(1+tanA)

=cotA / (1 + tanA)

= RHS

⟹ Hope it helps you :)