dont know how to solve (cosx+1)/(cotx+cscx)=sinx help please?

Question

any restrictions on the domain at all?

Answer 1

Is this a proof wanted or to solve the equation?
rewrite cotx as cosx/sinx and cscx as 1/sinx and you get

(cosx+1)/( cosx/sinx + 1/sinx) = sin x

sinx(cosx+1)/(cosx+1) = sinx QED

Answer 2

OK - how about this:

cotX = cosX/sinX cscX = 1/sinX

which means that you now could rewrite as

(cosX + 1) / [(cosX/sinX)+ (1/sinX)]
(cosX + 1) / [(cosX + 1) / sinX]

the cosX + 1 would cancel each other, resulting in sinX.

As for restrictions, any restriction placed on cot and csc functions would apply here. Check for restrictions for theses 2 functions.

Answer 3

Left hand factor: hit upon person-friendly denominator: ((a million+sinx)^2 + cos^2x) / (a million+sinx)(cosx) Foil out: (a million + 2sinx + sin^2x + cos^2x) / (a million+sinx)(cosx) bear in mind sin^2x + cos^2x = a million... (a million + 2sinx + a million) / (a million+sinx)(cosx) (2 + 2sinx) / (a million+sinx)(cosx) 2(a million + sinx) / (a million+sinx)cosx 2 / cosx = 2secx

Answer 4

(cotx+cscx) = cos/sinx + 1/sinx = (cosx + 1 ) / sinx
then (cosx+1)/(cotx+cscx)=sinx
Domain is R - {0 +/- 180n} ,n is an integral number

Answer 5

(cosx+1)/(cotx+cscx)=sinx
=>(1+cosx)/(cosx/sinx+1/sinx)=sinx
=>(1+cosx)/(1+cosx/sinx)=sinx
=>sinx=sinx
from here u can't find x
if r.h.s is sinA
then x=n(pi)(+-)x

Answer 6

LHS
(cos x + 1) / (cos x/sin x + 1 / sin x)
= sinx.(cosx + 1) / (cosx + 1)
= sin x

RHS
= sin x

LHS = RHS as required.