Trigonometric Identities: Prove (1+cosA-sinA)/(1+cosA+sinA)=secA-tanA
2 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
You need to multiply by `1+cosA+sinA` both sides such that:
`(1+cosA-sinA) = (1+cosA+sinA)(sec A - tan A)`
You need to substitute `1/cos A` for sec A and `sin A/cos A` for tan A such that:
`(1+cosA-sinA) = (1+cosA+sinA)(1/cos A-sin A/cos A)`
`(1+cosA-sinA) = (1+cosA+sinA)(1 -sin A)/cos A `
You need to multiply by cos A both sides such that:
`cos A*(1+cosA-sinA) = (1+cosA+sinA)(1 - sin A)`
You need to open the brackets such that:
`cos A + cos^2 A - cos A*sin A = 1 -sin A + cos A - cos A*sin A+ sin A - sin^2 A`
You need to reduce like terms such that:
`cos^2 A = 1 - sin^2 A`
`cos^2 A + sin^2A = 1`
The last line represents the basic formula of trigonometry, hence, the identity `(1+cosA-sinA)/(1+cosA+sinA)=secA-tanA` is checked.
to prove (1+cosA - sinA)/ (1+cosA+ sinA) = secA - tanA
divide both numerator and denominator by cosA
(1+cosA - sin)/ cosA / (1+cosA+sinA) / cosA
seperate the denominator cosA for each term , let me show you the simplification of both the numerator and denominator seperately
numerator: (1+cosA - sinA)/ cosA= 1/cosA + cosA/cosA - sinA/cosA
= secA +1 - tanA .(as 1/cosA = secA, sinA/cosA= tanA)
= secA-tanA +1
similarly, denominator: (1+ cosA + sinA)/ cosA = 1/cosA +cosA/cosA + sinA/cosA = secA + 1 + tanA.
= secA+tanA + 1
After simplification now we have Nr. / Dr. = (secA-tanA +1) / (secA+tanA+1)..................................(1)
let 1 = sec^2A - tan^2A in the denominator of equation (1)
then equation (1) is (secA-tanA + 1) / ((secA +tanA)+ (sec^2A - tan^2A))
but sec^2A- tan^2A = (secA-tanA)(secA+tanA), from the formula "a^2-b^2 = (a+b)(a-b)"................................(I)
plug in the formula ,to get (secA-tanA+1) / ((secA+tanA)+ (secA + tanA)(secA- tanA)),
now take (secA+tanA) common from the denominator, we get
(secA- tanA +1) / (secA+tanA) ( 1+ secA - tanA)
= (secA-tanA+1) / (secA +tanA)( secA-tanA + 1)
we see that (secA-tanA +1) is the like term of both the Nr. and Dr., so cancel out that term.
we get, 1 / (secA + tanA),...................(2)
multiply and divide (2) by (secA - tanA)
(secA -tanA)/ (secA+tanA)(secA - tanA)
= (secA -tanA) / (sec^2A - Tan^2A) = (secA -tanA) / 1
= secA - tanA
(using formula (I) )
thus we get (1+cosa - sinA) / (1+cosa + sinA) = secA - tanA