Question

Establish
cscu-cotu=sinu/(1+cosu)

Answer 1

if you replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\) then the left hand side is \[\frac{1}{b}-\frac{a}{b}\]

Answer 2

subtracting gives you
\[\frac{1-a}{b}\] or
\[\frac{1-\cos(x)}{\sin(x)}\]

Answer 3

so I get
\[\frac{ cosu - \sin ^2u }{ cosusinu } = \frac{ sinu }{ 1+cosu }\]

Answer 4

now lets see if we can show
\[\frac{1-\cos(x)}{\sin(x)}=\frac{\sin(x)}{1+\cos(x)}\]

Answer 5

how did you get the left side?

Answer 6

in your subtraction you overcomplicated it
you can use just \(\sin(x)\) as our denominator

Answer 7

\[\frac{1}{\sin(x)}-\frac{\cos(x)}{\sin(x)}=\frac{1-\cos(x)}{\sin(x)}\]

Answer 8

oh I got it thanks!

Answer 9

now how to get \[\frac{1-\cos(x)}{\sin(x)}=\frac{\sin(x)}{1+\cos(x)}\]? one hint is that
\[\sin^2(x)+\cos^2(x)=1\] or
\[\sin^2(x)=1-\cos^2(x)\] factoring on the left gives
\[\sin^2(x)=(1+\cos(x))(1-\cos(x))\]

Answer 10

or if you like
\[\sin(x)\times \sin(x)=(1+\cos(x))(1-\cos(x))\] divide both sides by \(1+\cos(x)\) and divide by \(\sin(x)\) and you get it
\[\frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)}\]

Answer 11

i meant "factoring on the right"

Answer 12

so the answer is
\[\frac{ 1-\cos^2u }{ sinu(1+cosu) }\]

Answer 13

for the left side

Answer 14

hmm no

Answer 15

you good to this line \[\sin(x)\times \sin(x)=(1+\cos(x))(1-\cos(x))\]

Answer 16

why?

Answer 17

ok i think i have confused you
i am not starting with \[\frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)}\] but i am going to end with it

Answer 18

but if you want to start with that, you can "cross multiply" and see that it is true
that is not really a proof, but it might do

Answer 19

\[\sin^2(x)+\cos^2(x)=1\\
\sin^2(x)=1-\cos^2(x)\\
\sin(x)\times \sin(x)=(1+\cos(x))(1-\cos(x))\] you can get from that line to what you want to prove, by dividing both sides by \(\sin(x)\) and by \(1+\cos(x))\)

Answer 20

thanks !

Answer 21

yw
hope i did not confuse you too much