Question

Help me; I don't know where to start.
Prove:
1) cotx- (cos2x/sinx cosx) = tanx

2) Sinx/sin^2x+cos2x = secx/cotx

3) (cscx(cosx+sinx)^2)/1+sin2x = cotx/cscx

Answer 1

Let's rewrite the equation:
cotx - tanx = (cos2x/sinxcosx)

Let's look at (cos2x/sinxcosx)

Use the double angle formula: \[\cos2x = \cos ^{2}x - \sin ^{2}x\]
Split up the numerators while keeping a common denominator: \[[(\cos ^{2}x)/(sinxcosx)] - (\sin ^{2}x)/(sinxcosx)]\]
After canceling we are left with: (cosx/sinx) - (sinx/cosx)

Use the reciprocal identities: cosx/sinx = cotx and sinx/cosx = tanx

So, this reduces to cotx-tanx

Answer 2

Hope this helps

Answer 3

oh...wait. I get it. thank you.

Answer 4

If you need help with the other two, let me know

Answer 5

are we allowed to rewrite the equation to cotx - tanx = (cos2x/sinxcosx) ?

Answer 6

yes, of course!

Answer 7

oh really? I was never told that... o_o

Answer 8

Do you think you could help me with (cscx(cosx+sinx)^2)/1+sin2x = cotx/cscx I think I got the 2nd problem but this one's confusing

Answer 9

simplify (cosx=sinx)^2 first. The rest will follow.

Answer 10

should read (cosx+sinx)^2

Answer 11

so it'll be cos^2x+2cosxsinx+sin^2x?

Answer 12

Yes. now what is cos^2x+sin^2x ?

Answer 13

1

Answer 14

so what do you have?

Answer 15

1+2cosxsinx

Answer 16

Next what is sin2x ?

Answer 17

1-cos^2x

Answer 18

wait opps...nvm

Answer 19

it's 2sinxcosx

Answer 20

correct

Answer 21

Can you simplify the whole equation now?

Answer 22

so do they cross out and leave cscx? or....

Answer 23

because I dont think they'll cross out since cscx is multiplying it

Answer 24

just write down the equation first

Answer 25

cscx(1+2sinxcosx)/1+2sinxcosx --> cscx+2sinxcosxcscx/1+2sinxcosx?

Answer 26

they will cancel out. what you will have left is cscx on the left hand side

Answer 27

Now simplify secx/cotx

Answer 28

Remember secx=(1/cosx) and cotx = (cosx)/(sinx)

Answer 29

where did you get the sec from?

Answer 30

from the right hand side of your original equation

Answer 31

there's no sec

Answer 32

sorry, i was looking at #2

Answer 33

it's cotx/cscx on the right side

Answer 34

cotx/cscx

Answer 35

so how does cscx alone get to cotx/cscx?

Answer 36

hmm! let me take a look again

Answer 37

give me a few minutes

Answer 38

I gotta go but try simplifying the equations again. We might have missed something.