Question

Prove:
\[\LARGE \frac{cotx}{cscx}=cosx\]

Answer 1

use this identities \[\cot x=\frac{\cos x}{\sin x}\]
\[\csc x=\frac{1}{\sin x}\] the rest is substitution

Answer 2

@Luigi210 and @Luigi0210 hahahaha, 2 of you.

Answer 3

ry expanding the left, and converting everything to sines and cosines.

(csc(x) - cot(x)) (csc(x) - cot(x)) =
csc^2(x) - 2cot(x)csc(x) + cot^2(x) =
(1/sin^2(x)) - 2(cos(x)/sin(x))(1/sin(x)) + (cos^2(x) / sin^2(x)) =
(1 - 2cos(x) + cos^2(x)) / sin^2(x) =
(1 - cos(x))^2 / sin^2(x) =
(1 - cos(x))^2 / (1 - cos^2(x)) =
(1 - cos(x))^2 / (1-cos(x))(1+cos(x)) =
(1 - cos(x)) / (1+cos(x))

Answer 4

So:
\[\Huge \frac{(\frac{cosx}{sinx})}{(\frac{1}{sinx})}=cosx\]
\[\Huge \frac{cosx}{sinx}*\frac{sinx}{1}=cosx\]
\[\Huge \frac{cosx}{1}*\frac{1}{1}=cosx\]
and finally:
\[\Huge cosx=cosx\]

@Loser66 Yea, someone is trying to be me I guess ._.

Answer 5

no u r trying 2 be me loser

Answer 6

wonder why there isn't anyone want to be me, a LOSER hehehehe

Answer 7

loooool Loser u shud change to Winner...cos nobody wants to be a looser

Answer 8

i wonder if @Luigi210 will also update her picture to look like thenew @Luigi0210

Answer 9

@Jonask , that makes me unique hehehe... glad to be a LOOOOOOser