Question

(1+cos alpha)/(sin alpha) +(sin alpha)/(1+cos alpha)
Simplify

Answer 1

\[\frac{ 1+\cos \alpha }{ \sin \alpha }+\frac{ \sin \alpha }{ 1+\cos \alpha}\]

Answer 2

I got it down to \[\frac{ 2 }{ \sin \alpha(1+\cos \alpha) }\]

Answer 3

Not sure thats right but >.<

Answer 4

what if you try multiplying top and bottom by the conjugate of \((1+\cos \alpha)\)?

Answer 5

k so that would end up like 2/sin alpha

Answer 6

which would be 2csc x

Answer 7

which is an answer.

Answer 8

and holy crap its right too! :O You're magic.

Answer 9

OKAY problem solved, onto the next one. I'll try to figure it out before i post it again.

Answer 10

multiplying by the conjugate is another trick worth remembering...

Answer 11

it's the (x-y)(x+y) = (x^2-y^2) bit all over again

Answer 12

okay. will try to remember/ understand.

Answer 13

It pops up in lots of places:
Simplifying radicals
\[\frac{1}{1+\sqrt{2}} = \frac{1}{1+\sqrt{2}}*\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{1-\sqrt{2}}{1^2-(\sqrt{2})^2} = \frac{1-\sqrt{2}}{-1} = \sqrt{2}-1\]
Complex numbers
\[\frac{1}{a+bi} = \frac{1}{a+bi}*\frac{a-bi}{a-bi} =\frac{a-bi}{a^2-b^2i^2} =\frac{a-bi}{a^2+b^2}\]
and so on

Answer 14

I may have to send you a gift basket after tonight my friend.

Answer 15

If you get the hang of doing this, that will be reward enough...but next time you have some pizza, eat an extra slice for me!

Answer 16

:) I just ate pretty much a whole homemade calzone in stress.