Question

Prove that :
CosA-sinA+1/cosA+sinA-1=secA+tanA

Answer 1

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Question:

Prove that:
(CosA-sinA+1) / (cosA+sinA-1) = secA+tanA

Answer:

To prove
(CosA-sinA+1) / (cosA+sinA-1) = secA+tanA

Divide both numerator and denominator by cosA

【(CosA-sinA+1) / cosA】 / 【(cosA+sinA-1) / 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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Answer 2

Step-by-step explanation:

Answer is in the pic.

Hope it will help you !