(1+sinA-cosA)/1+sinA+cosA=(1+sinA+c osA)/1+sinA-cosA=2cosecA

Hi,

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

sec2A=1+tan2Asec2A=1+tan2A

sec2A−tan2A=1sec2A−tan2A=1

Using above relation at denominator of LHS

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

Hence Proved.

I think above proof will clear your doubt,

All the best.

answer: what in the world is this

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