Prove that
sinA - cosA + 1/cosA + sinA -1 = 1/secA– tanA
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Answer:
sinA - cosA + 1/cosA + sinA -1 = 1/secA– tanA
LHS= sinA - cosA + 1 / cosA + sinA -1
dividing numerator and denominator by cosA
tanA -1+secA/1+tanA-secA =
=tanA+secA-1 /tanA-secA+1
=tanA+secA-(sec²A-tan²A)/tanA-secA+1
=tanA+secA-{(secA-tanA)(secA+tanA)}/tanA-sec+1
=tanA+secA{1-secA+tanA}/tanA-secA+1
=tanA+secA=sinA/cosA +1/cos A
=1+sinA/cosA × 1-sinA/1-sinA=1-sin²A/cosA(1-sinA)
=cos²A/cosA(1-sinA)= cosA/1-sinA
RHS = 1/secA-tanA = 1/1-sinA/cosA = cosA /1-sinA
LHS=RHS = cosA /1-sinA
hence proved
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