Question

how to prove this ? Question iv

Answer 1

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Answer 2

3)
LHS = cot²A/(cosecA-1)²

LHS = (cosec²A - 1)/(cosecA-1)(cosecA-1)

LHS = (cosecA+1)(cosecA-1)/(cosecA-1)(cosecA-1)

LHS = cosecA+1/cosecA-1

LHS = (1/sinA + 1)/(1/sinA -1)

LHS = (1+sinA/sinA)/(1-sinA/sinA)

LHS = 1+sinA/sinA × sinA/1-sinA

LHS = 1+sinA/1-sinA

LHS = RHS

4)
LHS = 1-(sin²A/1+cosA )

LHS = 1-(1-cos²A/1+cosA)

LHS = 1 - (1+cosA)(1-cosA)/(1+cosA)

LHS = 1 - (1-cosA)

LHS = 1-1+cosA

LHS = cosA

LHS = RHS

1)
LHS = 1+tan²A/1+cot²A

LHS = (1 + sin²A/cos²A)/(1 + cos²A/sin²A)

LHS = (cos²A+sin²A/cos²A)/(sin²A+cos²A/sin²A)

LHS = (1/cos²A)/(1/sin²A)

LHS = 1/cos²A × sin²A/1

LHS = sin²A/cos²

LHS = tan²A _____________(1)

RHS = (1-tanA)²/(1-cot)²
RHS = (1 - tanA/1-cotA)²

RHS = [(1 - sinA/cosA)/(1 - cosA/sinA)]²

RHS = [(cosA-sinA/cosA)/(sinA-cosA/sinA)]²

RHS = [(cosA-sinA/cosA) × (sinA/sinA-cosA)]²

RHS = [ (cosA-sinA/cosA) × sinA/-(cosA-sinA)]²

RHS = [ 1/cosA × sinA/-1 ]²

RHS = [ -sinA/cosA]²

RHS = (-tanA)² (-tanA × -tanA = +tan²A)

RHS = tan²A. _____________(2)

by comparing equation (1) and (2) we can say that
LHS = RHS