Question

1+tan^2A/ 1-cot^2A = [1-tanA/1-cotA] ^2

Answer 1

Step-by-step explanation:

bro i think there should be 1+cot^2A in question.

LHS =[1+tan^2A/1+cot^2A] = sec^2A/cosec^2A

[tanA = sinA/ cosA] and cotA = cosA / sinA] and [secA = 1/cosA] and cosecA= 1/sinA]

so , [1+(sin^2A/cos ^2A)/1+(cos^2A/sin^2A)] = (1/cos^2A)/1/sin^2A)

then, [( cos^A + sin^A)/cos^2A]/[(sin^2A+cos^2A)/sin^A] = (1/ cos^2A)/(1/sin^2A)

therefore ,(1/cos^2A)/(1+sin^2A) = (1/cos^2A)/(1/sin^2A)

therefore, (1/ cos^2A)/(1/sin ^2A) -sin^2A/cos^A = tan^2 A. LHS

RHS = [(1- tanA)/(1-cotA)] ^2 - [(1- tanA)/{1-(1/ tanA)}]^2

therefore,[{(1- tanA)/ tanA-1}/ tanA] ^2 - [(1- tanA)/( tanA- 1)×tanA]^2 = ( -tanA)^2 = tan^2A. RHS

so LHS= RHS