# 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.

justaguide | College Teacher | (Level 2) Distinguished Educator

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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

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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

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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).