Question

prove that sinA-cosA+1/sinA+cosA-1=1/secA-TanA​

Answer 1

Answer:

How do I prove sinA-cosA+1/sinA+cosA-1=1/secA-tanA?

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.

Answer 2

Answer:

Mark it as brainliest

Step-by-step explanation:

SinA-CosA+1/SinA+CosA-1

Divide num and den by cosA

TanA+SecA-1/TanA-secA+1

TanA+Seca-1/tan-seca+sec²a-tan²a{sec²a-tan²a=1}

TanA+Seca-1/tan-seca+(seca-tana)(seca+tana)

TanA+Seca-1/seca-tana(seca+tana-1)

TanA+Seca-1/seca-tana(seca+tana-1)

1/seca-tana