prove that SinA-cosA+1/sina A + Cos A-1 = 1/sec A - Tan A
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Answer:
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.
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