Solve this equation algebraically:
(1-sin x)/cos x = cos x/(1+sin x)
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I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but I still feel stuck. This is what I have so far:
cos^2(x)/cos x + sin x =
cos x - sin x/cos^2(x)
...but I'm not really sure how to get to the answer. Help please?
Thank you!
Do you mean solve it or prove it?
It is an identity, so there really isn't a specific solution: it's true for all x.
I suggest you try reformatting as
(1-sin x)/cos x - cos x/(1+sin x) = 0
Then bring both fractions to the common denominator (cosx)(1+sinx), and I think you'll recognise the numerator you're left with!
Ok, I worked it out, and so far i have this:
cos^2x - cos x - sinxcosx /
cosx + sinxcosx
Now I just need some help with reducing?
Um, no. Wrong turn somewhere. It's a LOT simpler than that.
Your numerator will be :
(1-sinx)(1+sinx) - cos^2x
To solve the equation algebraically, we can start by multiplying both sides of the equation by (1+sin x) and cos x to get rid of the denominators:
(1-sin x)(1+sin x) = cos^2 x
Expanding and simplifying the left side of the equation:
1 - sin^2 x = cos^2 x
Using the Pythagorean identity sin^2 x + cos^2 x = 1, we can rewrite the equation as:
1 - sin^2 x = 1 - cos^2 x
Now, we can subtract 1 from both sides of the equation:
- sin^2 x = - cos^2 x
Finally, we can divide both sides of the equation by -1 to simplify:
sin^2 x = cos^2 x
Using the Pythagorean identity cos^2 x = 1 - sin^2 x, we can rewrite the equation as:
sin^2 x = 1 - sin^2 x
Adding sin^2 x to both sides of the equation:
2sin^2 x = 1
Dividing both sides of the equation by 2:
sin^2 x = 1/2
Taking the square root of both sides of the equation:
sin x = ±√(1/2)
Using the values for sin x from the unit circle, we can find the solutions to the equation:
x = π/4 + nπ/2, where n is an integer
So, the solution to the equation algebraically is x = π/4 + nπ/2, where n is an integer.