Here's the question you clicked on:
bridgetx516x
Simplify the expression as much as possible: cscx / tanx+cotx
\[\frac{ cosec(x) }{ \tan(x)+\cot(x) }= \frac{ \frac{ 1 }{ \sin(x) } }{ \frac{ \sin(x) }{ \cos(x) }+\frac{ \cos(x) }{ \sin(x) } }\] So if we move the sin(x) to the bottom like so \[= \frac{ 1 }{ \sin(x)\frac{ \sin(x) }{ \cos(x) }+ \sin(x)\frac{ \cos(x) }{ \sin(x) } }\] Expand out the bottom further to get \[= \frac{ 1 }{ \frac{ \sin^2(x) }{ \cos(x) }+\cos(x) }\] Now let cos(x)=cos^2(x)/cos(x) and take out a factor of 1/cos(x) on the bottom.\[=\frac{ 1 }{ \frac{ \sin^2(x)+\cos^2(x) }{ \cos(x) } }\]As we know, sin^2(x) + cos^2(x) = 1, so we are left with 1/(1/cos(x)) which is just \[\frac{ cosec(x) }{ \tan(x)+\cot(x) }= \cos(x)\]Hope this helped:)
Paddy, you should have gotten a medal or at least a thank you for all that work!
There, I gave you a medal.
THANK YOU SO MUCH!