Question

(cosx+sinx)(cscx-secx)=cotx-tanx

Answer 1

are we doing identities :)

Answer 2

or are we solving for x?

Answer 3

or are can we do anything with it?

Answer 4

identities

Answer 5

fun!

Answer 6

Foil it out giving:
cos(x)csc(x)-sin(x)sec(x)+sin(x)csc(x)-sec(x)cos(x)=cot(x)-tan(x)
csc(x)=1/sin(x)
sec(x)=1/cos(x)
so you have:
cos(x)/sin(x)-sin(x)/cos(x)+sin(x)/sin(x)-cos(x)/cos(x)=cot(x)-tan(x)
cot(x)-tan(x)+1-1=cot(x)-tan(x)
Done.

Answer 7

Haha :P

Answer 8

DING! LEVEL UP!

Answer 9

we shall race again!

Answer 10

(cosx+sinx)(cscx-secx)=cotx-tanx
cosx(cscx)+cosx(cscx)-secx(cosx)+sinx(cscx)-sinx(secx)
csc(x)=1/sin(x)
sec(x)=1/cos(x)
cos(x)/sin(x)-sin(x)/cos(x)+sin(x)/sin(x)-cos(x)/cos(x)
cotx-tanx-1+1=cotx-tanx

Answer 11

sorry didnt saw ur reply

Answer 12

what are the domain restrictions?

Answer 13

cotx=cosx/sinx
and
tanx=sinx/cosx
make sure these bottoms aren't zero

Answer 14

For cotangent:
\[x \ne n \pi, n \in \mathbb{Z}\]
For tangent:
\[x \ne \frac{(2n+1)\pi}{2},n \in \mathbb{Z}\]

Answer 15

sinx is zero when x=0,pi,2pi, and so on...
in general sinx is zero when x=2pi*n for n=...-2,-1,0,1,2,3...
and sinx is zero when x=pi+2n*pi for n=...-2,-1,0,1,2,3,...
cosx is zero when x=pi/2 and 3pi/2
in general cosx is zero when x=pi/2+2pi*n and x=3pi/2+2pi*n
for n=integer
so the domain is all numbers not including any of the ones in mentioned above

Answer 16

I WIN AGAIN. HAHAHHAHAHAH.

Answer 17

let me say all real numbers not including the ones i just mentioned lol

Answer 18

very good
i have taught u well

Answer 19

Haha xP

Answer 20

\[\huge\int\limits IDENTITIES*dBRAINPOWER\]