How do you prove this equation?

Question

1+cos(x)/sin(x) + sin(x)/1+cos(x) = 2csc(x)

I just don't understand why it would turn out like that...so if someone could please help me out here, it is totally frying my mind. Thanks

Answer 1

To understand you should read over this section on Trigonometry identities on the site www.wikipedia.org and form here you will be able to reagrane the equation since somethings equal others

Answer 2

start with just one side:1+cos(x)/sin(x) + sin(x)/(1 + cos(x))

combine the terms: ((1+cos(x)^2)+sin(x)^2)/(sin(x)(1+cos(x))

simplify: (1+2cos(x)+cos(x)^2+sin(x)^2)/(sin(x)(1+cos(x))

cos(x)^2+sin(x)^2)=1 so:(2+2cos(x))/(sin(x)(1+cos(x))

factoring out the 2: 2(1+cos(x))/(sin(x)(1+cos(x))

the 1+cos(x) cancels: 2/sin(x)=2csc(x)

Answer 3

Your question is wrong:

put x=pi/4

=>LHS=1+[(1/sqrt(2))/(1/sqrt(2))] + (1/sqrt(2))/(1+1/sqrt(2))
=1+1+ 1/(sqrt(2) + 1)=2+(sqrt(2) - 1)

now, RHS=2*sqrt(2)
clearly LHS=/=RHS

also
LHS

1+cos(x)/sin(x) + sin(x)/(1+cos(x) )= 2csc(x)
LHS=1+cos(x)/sin(x) + sin(x)/1+cos(x)
take LCM

{sin(x)[1+cos(x)] + cos(x)[1+cos(x)] + sin(x)sin(x)}/sin(x)(1+cos(x)}

=>{sin(x) + sin(x)cos(x) + cos(x) + [cos(x)]^2 + [sin(x)]^2}/sin(x)[1+cos(x)]

=>{sin(x) + sin(x)cos(x) + 1 +cos(x)}/sin(x)[1+cos(x)]....................
........................since [cos(x)]^2 + [sin(x)]^2 =1

=>{sin(x)[1+cos(x)] + [1+cos(x)]}/sin(x)[1+cos(x)]

=>{[1+cos(x)][sin(x) + 1]}/sin(x)[1+cos(x)]......taking [1+cos(x)] comman in the numerator

=>[1+sin(x)]/sin(x)
=>cosec(x) + 1=/= 2cosec(x)


LHS=left hand side
RHS=right hand side

Answer 4

looks like you are doing your homework!

here is the trick, dont post one question more than once, that shows you really need the answer, which may result in not getting any answer at all.

Answer 5

wouldnt have a clue