trig nightmare: solve cosx/1 - sinx
- Thread starter seekingh
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Re: restate problem
. . . . .\(\displaystyle \L \frac{\cos{(x)}}{1}\,-\,\sin{(x)}\)
. . . . .\(\displaystyle \L \cos{\left(\frac{x}{1}\right)}\,-\,\sin{(x)}\)
Is either of these what you meant?
When you reply with confirmation or correction, please include everything you have tried thus far. Thank you.
Eliz.
What you have posted means one of the following:seekingh said:
. . . . .\(\displaystyle \L \frac{\cos{(x)}}{1}\,-\,\sin{(x)}\)
. . . . .\(\displaystyle \L \cos{\left(\frac{x}{1}\right)}\,-\,\sin{(x)}\)
Is either of these what you meant?
When you reply with confirmation or correction, please include everything you have tried thus far. Thank you.
Eliz.
no, neith er one is correct. the expression is wriiten cos x as the numerator and 1-sin x as the denominator. I have tried using the basic formulas of y/r for sin and x/r for cos. However I getting sec x+ tan x, which is not a correct answer according to my book
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So you mean your expression to be as follows...?seekingh said:
. . . . .\(\displaystyle \L \frac{\cos{(x)}}{1\,-\,\sin{(x)}}\)
What have you done? What steps did you take?seekingh said:
This expression is already fairly simple. It is difficult to say what the book might have in mind. If you have "the" answer, it might help if you provided that, as this could guide our advice to you.seekingh said:
Thank you.
Eliz.
skeeter
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\(\displaystyle (\frac{cosx}{1-sinx})(\frac{1+sinx}{1+sinx}) =\)
\(\displaystyle \frac{cosx(1+sinx)}{1-sin^2x} =\)
\(\displaystyle \frac{cosx(1+sinx)}{cos^2x} =\)
\(\displaystyle \frac{1+sinx}{cosx}\)
note that the last expression also equals \(\displaystyle secx + tanx\), which is what you stated earlier ... so, what did the "book" have as an answer?
\(\displaystyle \frac{cosx(1+sinx)}{1-sin^2x} =\)
\(\displaystyle \frac{cosx(1+sinx)}{cos^2x} =\)
\(\displaystyle \frac{1+sinx}{cosx}\)
note that the last expression also equals \(\displaystyle secx + tanx\), which is what you stated earlier ... so, what did the "book" have as an answer?