Prove that (sin A - cos A + 1)/(sin A + cos A - 1) =cos A/1-sin ATHIS QUESTION IS FROM CHAPTER TRIGNOMETRIC IDENTITIES AND IN THIS QUESTION WE HAVE TO PROVE THE LEFT HAND SIDE=RIGHT HAND SIDE.
3 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to prove that (sin A - cos A + 1)/(sin A + cos A - 1) = cos A/(1 - sin A)
Start from the left hand side
(sin A - cos A + 1)/(sin A + cos A - 1)
=> (sin A - cos A + 1)(sin A - cos A -1)/(sin A + cos A - 1)(sin A - cos A -1)
=> ((sin A - cos A)^2 - 1)/((sin A - 1)^2 - (cos A)^2)
=> ((sin A - cos A)^2 - 1)/((sin A)^2 - 2*sin A + 1 - (cos A)^2)
=> ((sin A - cos A)^2 - 1)/((sin A)^2 - 2*sin A + (sin A)^2)
=> ((sin A)^2 + (cos A)^2 - 2*sin A*cos A - 1)/((sin A)^2 - 2*sin A + (sin A)^2)
=> (1- 2*sin A*cos A - 1)/((sin A)^2 - 2*sin A + (sin A)^2)
=> (-2*sin A*cos A)/(2*(sin A)^2 - 2*sin A)
=> (-cos A)/(sin A - 1)
=> cos A/(1 - sin A)
which is the right hand side
This proves:(sin A - cos A + 1)/(sin A + cos A - 1) = cos A/(1 - sin A)
lochana2500 | Student, Undergraduate | (Level 1) Valedictorian
L:H:S = (sinA-cosA+1)/(sinA+cosA-1)
devide the numerator and denominator by cosA
= (tanA-1+secA)/(tanA+1-secA)
= (secA+tanA-1)/(1-secA+tanA)
= [secA+tanA-(sec²A-tan²A)]/(1-secA+tanA)
=[secA+tanA-(secA+tanA)(secA-tanA)/(1-secA+tanA)
= (secA+tanA)(1-secA+tanA)/(1-secA+tanA)
= secA+tanA
= (1+sinA)/cosA
multiply the numerator and denominator by (1-sinA)
= (1-sin²A)/cosA(1-sinA)
= cos²A/cosA(1-sinA)
= cosA/(1-sinA)
= R:H:S
neela | High School Teacher | (Level 3) Valedictorian
To Prove that (sin A - cos A + 1)/(sin A + cos A - 1) = cos A/(1-sin A)
We multiply both numerator and denominator of LHS by (sinA-cosA-1):
(sin A - cos A + 1)(sinA-cosA-1)/(sin A + cos A - 1)(sinA-cosA-1)
={(sinA-cosA)^2-1}/{{(sinA-1)^2-cos^2 A)}, as (a+b)(a-b) = a^2-b^2.
={sin^2A+cos^2A-2sinAcosA-1}/{sin^2A+1-2sinA-1+sin^2 A}}, as sin^2x+cos^2x = 1.
= -2sinAcosA/2sinA(sinA-1)
= cosA/(1-sinA) = RHS.
Therefore (sin A - cos A + 1)/(sin A + cos A - 1) =cos A/(1-sin A).