Trigonometric Identities: 
Prove (1+cosA-sinA)/(1+cosA+sinA)=secA-tanA

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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.

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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

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