Question

Simplify the following. Your answer should only involve the cosine function.
(cosx/1-sinx)-tanx

Answer 1

\[\frac{a}{1-b}-\frac{b}{a}\] is the first step

Answer 2

anyway u can go more in depth? i wasnt their when we learned it so i really dont get it...U can give me the answer as long as you explain how it is you got it..please and thank u

Answer 3

most of it is algebra, don't let the sine and cosine confuse you, it has very little to do with the problem
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\] so you have
\[\frac{\cos(x)}{1-\sin(x)}-\frac{\sin(x)}{\cos(x)}\]too much writing
if it put \(a=\cos(x), b=\sin(x)\) then the algebra is clearer, you get
\[\frac{a}{1-b}-\frac{b}{a}\] and you can subract (algebra for this)

Answer 4

subtraction give you
\[\frac{a}{1-b}-\frac{b}{a}=\frac{a^2-b(1-b)}{(1-b)a}=\frac{a^2+b^2-b}{(1-b)a}\]

Answer 5

now back to trig,
\[\frac{\cos^2(x)+\sin^2(x)-\sin(x)}{(1-\sin(x))\cos(x)}\] and since \(\cos^2(x)+\sin^2(x)=1\) you get
\[\frac{1-\sin(x)}{(1-\sin(x))\cos(x)}\] that was the only trig involved, the rest was algebra alone

Answer 6

last step is to cancel, and you are done

Answer 7

so it would be 1/cosx ?

Answer 8

omg...thanks alot i get it..when over it a couple time but i understand what you mean know thanks