Can anyone explain how to prove this using trigometric identidies?

Question

anonymous · Answer

(sec(x)+csc(x))/(1+tan(x)) = csc(x)
prove it by manipulating only the left side of the equation

superdavesuper · Answer

easiest way is probably start by expressing everything on the left in terms of sin(x) and cos(x)....what will u get?

anonymous · Answer

(1/cos(x))+(1/sin(x))

superdavesuper · Answer

yup yup....that would be for the top part....how about the bottom? tanx=sinx/cosx right? :)

anonymous · Answer

yeah, so then do you multiply the top like this?
(1/cos(x))(cos(x)/sin(x)) + (1/sin(x))(cos(x)/sin(x))

superdavesuper · Answer

nah i would just multiply both top n bottom by (sinx)(cosx)....that will get rid of the division in the top first...

anonymous · Answer

???

superdavesuper · Answer

(1/cos(x))(cos(x)(sin(x)) + (1/sin(x))(cos(x)(sin(x)) = sin(x) + cos(x) right? need to do the same multiplication to the bottom part though

anonymous · Answer

how do you do this while only manipulating the left side?

superdavesuper · Answer

by multiplying the same thing (cosx)(sinx) to BOTH top n bottom on the left

anonymous · Answer

so...
((1/cosx)(cosx)(sinx)+(1/sinx)(cosx)(sinx))/(1(sinx)(cosx)+((sinx)/(cosx))(sinx)(cosx)???

superdavesuper · Answer

hahahahahahahaa yes - i would probably have splitted the top n bottom up n do them separately but u got it :)

anonymous · Answer

does it become...
((((sinx)(cosx))/cosx)+(1/sinx)(sinx)(cosx))/(sinx)(cosx)+sin^2x

superdavesuper · Answer

lets do it separately okay? im getting a head ache from reading lol

top part is done: it would simplify to sinx+cosx okay?

superdavesuper · Answer

n u got the bottom part right (sinx)(cosx) + (sinx)^2

superdavesuper · Answer

see anyhting similar between top n bottom? ;)

anonymous · Answer

okay

anonymous · Answer

um the bottom becomes (sinx)(cosx+sinx) and that cancels out with the the sinx+cosx on the top

anonymous · Answer

does that mean your left with 1/sinx which equals cscx?

superdavesuper · Answer

Bingo :)

anonymous · Answer

thank you!!!

superdavesuper · Answer

welcome - u did most of the work anyway lol

anonymous · Answer

i have the most troble with trig proofs so thank you soo much!!!

superdavesuper · Answer

n thanx for staying w the prob. a lot of ppl would simply ask for the proof itself :)