Question

sinA-cosA+1÷sinA+cosA-1=1÷secA-tanA
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Answer 1

Answer:

Hi,

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

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)

Now

sec2A=1+tan2A

sec2A−tan2A=1

Using above relation at denominator of LHS

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

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

LHS=1/(secA−tanA)

LHS=RHS

Hence Proved.

I think above proof will clear your doubt,

All the best.

Answer 2

Answer in attachment.

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