can someone help? ( cos x - sec x /sec x +cos^2 x tan^2 x)(tan x -sin x / tan x) I need to express this in term of sine and cosine and simplify. spent like 3 hrs doing this and still can't get a correct answer what I get is (0)(-sin x) which I know is not a right answer.
trig question please help
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Harry_the_cat
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Is this what you mean?
\(\displaystyle \frac{cosx - secx}{secx + cos^2x tan^2x}*\frac{tanx-sinx}{tanx}\)
\(\displaystyle \frac{cosx - secx}{secx + cos^2x tan^2x}*\frac{tanx-sinx}{tanx}\)
Dr.Peterson
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Please, please, please use parentheses appropriately! It saves a lot of time when you say exactly what you mean!
The function is apparently this:
(((cos x - sec x) /sec x) + cos^2 x tan^2 x)((tan x -sin x) / tan x)
which you have not yet written clearly. It means this:
[MATH]\left(\frac{cosx - secx}{secx} + cos^2x tan^2x\right)*\frac{tanx-sinx}{tanx}[/MATH]
When I simplify this, I get 0, as you did (except that you didn't carry out the multiplication). How do you know that is incorrect?
Yes sorry that is the function. So then when I Express in terms of sine and cosine I am left with [MATH](cos^2x-1+sin^2x)[/MATH] which simplifies to just 0 but the other part when I Express it in terms of cosine and sin I get[MATH]((sinx/cosx)-sinx)/(sinx/cosx)[/MATH] so if I do devision I am just left with - sinx but if I start subtracting numorator first I get different answer, what is a right way to do?
Dr.Peterson
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No, the second factor does not reduce to -sin(x); I suspect you "canceled" illegally. Check what you did there.
But that doesn't matter (in terms of the answer), because when you multiply anything by zero, you get zero.
Please show your work for both of the answers you got, so we can see your errors.
Dr.Peterson
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As I suggested, you "canceled illegally" - on the very last line.
You can't divide just one term of the numerator by sin(a)/cos(a). If you're going to do that at all, you have to do it to both terms.
You might do this: [MATH]\left(\frac{\sin a}{\cos a} - \sin a\right) \div \frac{\sin a}{\cos a} = \left(\frac{\sin a}{\cos a} - \sin a\right) \times \frac{\cos a}{\sin a} = \frac{\sin a}{\cos a} \times \frac{\cos a}{\sin a} - \sin a \times \frac{\cos a}{\sin a} = 1 - \cos a[/MATH].
On the line beginning "or", you made a different mistake, apparently "subtracting" a factor and leaving 1. [MATH]\sin a - \sin a \cos a \ne \cos a[/MATH]; rather, [MATH]\sin a - \sin a \cos a = \sin a(1 - \cos a)[/MATH]. This will give you the same correct result I got the other way.
Watch your algebra when you do trigonometry!
You can't divide just one term of the numerator by sin(a)/cos(a). If you're going to do that at all, you have to do it to both terms.
You might do this: [MATH]\left(\frac{\sin a}{\cos a} - \sin a\right) \div \frac{\sin a}{\cos a} = \left(\frac{\sin a}{\cos a} - \sin a\right) \times \frac{\cos a}{\sin a} = \frac{\sin a}{\cos a} \times \frac{\cos a}{\sin a} - \sin a \times \frac{\cos a}{\sin a} = 1 - \cos a[/MATH].
On the line beginning "or", you made a different mistake, apparently "subtracting" a factor and leaving 1. [MATH]\sin a - \sin a \cos a \ne \cos a[/MATH]; rather, [MATH]\sin a - \sin a \cos a = \sin a(1 - \cos a)[/MATH]. This will give you the same correct result I got the other way.
Watch your algebra when you do trigonometry!
topsquark
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Just for consumption, I think you did write this correctly. The underlines here act like a pair of parenthesis, or more in line with what Mathematics calls a "vinculum," which is merely an overline as opposed to an underline. That being said it is non-standard.
-Dan