verifying [(1 + cosx)/sinx] + [sinx/(1 + cosx)] = 2cscx

[stkarnivor]

I need help understanding what I am missing

problem: ( ( 1 + cosx ) / sinx ) + ( sinx / ( 1 + cosx ) ) = 2cscx

i got as far as: 1 + cosx + sinx = 2cscx

 

[Loren]

Re: verifying trig identities

There is no one way to prove an identity. We need to see each of your steps if we are going to steer you in a right way.
I found it quite direct to add the two fractions on the left. In so doing you will end up with some squares. Expand the binomial and simplify. You should see it from there.

 

[stkarnivor]

Re: verifying trig identities

1. ( ( 1 + cosx ) / sinx ) + ( sinx / ( 1 + cosx ) ) = 2cscx -problem
2. ( ( 1 + cosx )^2 + ( sinx )^2 ) / ( sinx * ( 1 + cosx) ) = 2cscx -cross multiply
3. ( ( 1 + cosx ) + ( sinx ) ) / 1 = 2cscx -simplify
4. ( ( ^$#^#)($#$^(#$^)#^#^(@)*!($ )!!!!! = 2cscx -get stuck

 

[daon]

Re: verifying trig identities

1. ( ( 1 + cosx ) / sinx ) + ( sinx / ( 1 + cosx ) ) = 2cscx -problem
2. ( ( 1 + cosx )^2 + ( sinx )^2 ) / ( sinx * ( 1 + cosx) ) = 2cscx -cross multiply
3. ( ( 1 + cosx ) + ( sinx ) ) / 1 = 2cscx -simplify
4. ( ( ^$#^#)($#$^(#$^)#^#^(@)*!($ )!!!!! = 2cscx -get stuck

lol @ 4

Take 2. FOIL out the 1+cosx. The numerator becomes:

1 + 2cosx + (cosx)^2 + (sinx)^2 = 2(1+cosx)

Good?

 

[stkarnivor]

Re: verifying trig identities

Thanks for the help, i guess ill just have to brush up on my basic algebra skills