Sorry to ask again, but I need help with these problems:
(secx+1)/tanx = sinx/(1-cosx)
I got (1+cosx)/sinx which is the opposite of what I need.
tan^4x + tan^2x = sec^4 - sec^2x
tan^2x - sin^2x = tan^2x * sin^2x
sinx/(1+cosx) + (1+cosx)/sinx = 2cscx
I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring.
this works, but it's overkill. you could go directly from . just combining fractions, as we know . from there, it's nice
we could also do
@ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go.
Oh, right. Duh. I had that in my notes and still couldn't figure it out.
Yeah, but I usually pick the wrong ways to solve them.@ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go.
Thanks to everyone for all of their help!!
that's the point. as long as your moves are mathematically sound, and you can get from one side to the other, there is no wrong way. a problem here can be done several valid ways. the difference is, one way might be easier or quicker than another. if you do enough practice, this won't matter though, as you will either be able to move so quickly that the extra steps hardly slow you down, or you will naturally be able to spot the easier ways first