SOLUTION: Prove that : (sin x - cos x + 1)/(sin x + cos x -1) = 1 / (sec x - tan x) Prove that : (tan x + sec x -1)/(tan x - sec x +1) = (1+sin x)/ cos x

Algebra ->  Trigonometry-basics -> SOLUTION: Prove that : (sin x - cos x + 1)/(sin x + cos x -1) = 1 / (sec x - tan x) Prove that : (tan x + sec x -1)/(tan x - sec x +1) = (1+sin x)/ cos x       Log On


   

Question 878293: Prove that :
(sin x - cos x + 1)/(sin x + cos x -1) = 1 / (sec x - tan x)
Prove that :
(tan x + sec x -1)/(tan x - sec x +1) = (1+sin x)/ cos x
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
Prove

%28sin%28x%29-cos%28x%29+%2B+1%29%2F%28sin%28x%29%2Bcos%28x%29+-1%29%22%22=%22%22%28sin%28x%29-cos%28x%29+%2B+1%29%2F%28sin%28x%29%2Bcos%28x%29+-1%29

LEFT SIDE = %28sin%28x%29-cos%28x%29+%2B+1%29%2F%28sin%28x%29%2Bcos%28x%29+-1%29

RIGHT SIDE = 

1%22%F7%22%28sec%28x%29-tan%28x%29%29

1%22%F7%22%281%2Fcos%28x%29-sin%28x%29%2Fcos%28x%29%29

1%22%F7%22%281-sin%28x%29%29%2Fcos%28x%29

1%22%D7%22cos%28x%29%2F%281-sin%28x%29%29

cos%28x%29%2F%281-sin%28x%29%29

Now we need to show that

%28sin%28x%29-cos%28x%29+%2B+1%29%2F%28sin%28x%29%2Bcos%28x%29+-1%29%22%22=%22%22cos%28x%29%2F%281-sin%28x%29%29

Let's use S for sin(x) and C for cos(x)

%28S-C+%2B+1%29%2F%28S%2BC-1%29%22%22=%22%22C%2F%281-S%29

And work with the left side and show that it equals C%2F%281-S%29

%28%28S-C%29+%2B+1%29%2F%28%28S%2BC%29-1%29

Multiply top and bottom by (S-C)-1



%28%28S-C%29%5E2-1%29%2F%28%28S%2BC%29%28S-C%29-%28S%2BC%29-%28S-C%29%2B1%29+

%28S%5E2-2SC%2BC%5E2-1%29%2F%28S%5E2-C%5E2-S-C-S%2BC%2B1%29

%28S%5E2%2BC%5E2-1-2SC%29%2F%28S%5E2-C%5E2-2S%2B1%29

Since S%5E2%2BC%5E2=1 the numerator becomes just -2SC
Write the 1 in the denominator as S%5E2%2BC%5E2

%28-2SC%29%2F%28S%5E2-C%5E2-2S%2BS%5E2%2BC%5E2%29

%28-2SC%29%2F%282S%5E2-2S%29

%28-2SC%29%2F%28-2S%28-S%2B1%29%29

Divide top and bottom by -2S

C%2F%28-S%2B1%29 =

C%2F%281-S%29

--------------------------------------

Prove

%28tan%28x%29%2Bsec%28x%29+-+1%29%2F%28tan%28x%29-sec%28x%29%2B1%29%22%22=%22%22%281%2Bsin%28x%29%29+%2Fcos%28x%29%29

LEFT SIDE = %28tan%28x%29%2Bsec%28x%29+-+1%29%2F%28tan%28x%29-sec%28x%29%2B1%29

RIGHT SIDE = %281%2Bsin%28x%29%29+%2Fcos%28x%29%29

Work with the left side

%28tan%28x%29%2Bsec%28x%29+-+1%29%2F%28tan%28x%29-sec%28x%29%2B1%29 =

%28tan%28x%29%2Bsec%28x%29+-+1%29%22%F7%22%28tan%28x%29-sec%28x%29%2B1%29 =

%28sin%28x%29%2Fcos%28x%29%2B1%2Fcos%28x%29-1%29%22%F7%22%28sin%28x%29%2Fcos%28x%29-1%2Fcos%28x%29%2B1%29 =

%28+sin%28x%29%2B1-cos%28x%29%29%2Fcos%28x%29%29%22%F7%22%28sin%28x%29-1%2Bcos%28x%29%29%2Fcos%28x%29 =

%28+sin%28x%29%2B1-cos%28x%29%29%2Fcos%28x%29%29%22%D7%22cos%28x%29%2F%28sin%28x%29-1%2Bcos%28x%29%29 =


%28+sin%28x%29%2B1-cos%28x%29%29%2F%28sin%28x%29-1%2Bcos%28x%29%29

So the identity to prove is

%28+sin%28x%29%2B1-cos%28x%29%29%2F%28sin%28x%29-1%2Bcos%28x%29%29 = %281%2Bsin%28x%29%29+%2Fcos%28x%29%29

Using S for sin(x) and C for cos(x),

%28S%2B1-C%29%2F%28S-1%2BC%29 = %281%2BS%29%2FC

Working with the left side:

%28%28S%2B1%29-C%29%2F%28%28S-1%29%2BC%29 =

Multiply by %28%28S%2B1%29%2BC%29%2F%28%28S%2B1%29%2BC%29 which just equals 1:

%28%28S%2B1%29-C%29%2F%28%28S-1%29%2BC%29%22%D7%22%28%28S%2B1%29%2BC%29%2F%28%28S%2B1%29%2BC%29 =

 =

%28S%5E2%2B2S%2B1-C%5E2%29%2F%28S%5E2-1%2BCS-C%2BCS%2BC%2BC%5E2%29 =

%28S%5E2%2B2S%2B1-C%5E2%29%2F%28S%5E2%2BC%5E2-1%2B2CS%29 =

Since S%5E2%2BC%5E2=1, the denominator is just 2CS

%28S%5E2%2B2S%2B1-C%5E2%29%2F%282CS%29 =

Write the 1 in the numerator as S%5E2%2BC%5E2

%28S%5E2%2B2S%2BS%5E2%2BC%5E2-C%5E2%29%2F%282CS%29 =

%282S%5E2%2B2S%29%2F%282CS%29 =

%282S%28S%2B1%29%29%2F%282CS%29 =

Divide top and bottom by 2S

%28S%2B1%29%2FC =

%281%2BS%29%2FC

Edwin