Question

(sinx)/(cscx+cotx) Help! :) Simplify trig. expression.

Answer 1

sorry, I suck at trig. :/

Answer 2

\[sinx/cscx+cotx\]
\[\frac{ sinx }{ \frac{ 1 }{ sinx } \frac{ cosx }{ sinx }}\]

Answer 3

Thanks for trying, owl eyes :)

Answer 4

\[\begin{align*}\frac{\sin x}{\csc x+\cot x}&=\frac{\sin x}{\dfrac{1}{\sin x}+\dfrac{\cos x}{\sin x}}\times\frac{\sin x}{\sin x}\\\\&=\frac{\sin^2x}{1+\cos x}\\\\&=\frac{1-\cos^2x}{1+\cos x}\\\\&=\cdots\end{align*}
\]

Answer 5

Wait, on the second step how did you get to 1+cosx in the denominator?

Answer 6

I get the sin/sin = 1
but (cos/sinx) times sin?

Answer 7

@zepdrix ? :)

Answer 8

He multiplied top and bottom by sin x.\[\Large\rm \frac{\sin x}{\dfrac{1}{\sin x}+\dfrac{\cos x}{\sin x}}\color{orangered}{\left(\frac{\sin x}{\sin x}\right)}=\frac{\sin x\cdot\color{orangered}{\sin x}}{\dfrac{\color{orangered}{\sin x}}{\sin x}+\dfrac{\color{orangered}{\sin x}\cdot\cos x}{\sin x}}\]\[\Large\rm =\frac{\sin^2x}{1+\cos x}\]ya? :d

Answer 9

bahh i gotta go >.<

Answer 10

oou ok that makes sense. Thanks!

Answer 11

Well byee :P

Answer 12

Wait, now what does it equal?

Answer 13

sin^2x = what ?
do you know ?

Answer 14

yes ? no ? maybe ?? something ? nothing ?

Answer 15

sin^2x= 1-cos^2

Answer 16

hmm i guess that's it
sin^2 x/1+cos x

Answer 17

\[\frac{ 1-\cos^2x }{1+cosx}\]
now what? :P

Answer 18

i guess that's not gonna work
well i thought we can apply difference of square (a+b)(A-b) hmm but no iguess

Answer 19

i guess its suppose to be sinx+cosx at the denominator
hmm

Answer 20

gtg sorry
i willl solve it

Answer 21

okies. thanks for helping!

Answer 22

Factor as a difference of squares: \(a^2-b^2=(a-b)(a+b)\). Set \(a=1\) and \(b=\cos x\).