Question

Prove that sinA-cosA+1÷sinA+cosA-1 = 1÷secA-tanA​

Answer 1

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)

LHS=(tanA−1+secA)/(tanA+1−secA)

Now

se

c

2

A=1+ta

n

2

A

sec2A=1+tan2A

se

c

2

A−ta

n

2

A=1

sec2A−tan2A=1

Using above relation at denominator of LHS

LHS=(tanA−1+secA)/(tanA−secA+se

c

2

A−ta

n

2

A)

LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=1/(secA−tanA)

LHS=1/(secA−tanA)

LHS=RHS

LHS=RHS

Hence Proved.