Math - help really needed

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I'm sorry to double post; I don't want to seem impatient, but I really need help with this.

Prove each idenity.

1+1/tan^2x=1/sin^2x

1/cosx-cosx=sinxtanx

1/sin^2x+1/cos^2x=1/sin^2xcos^2x

1/1-cos^2x+/1+cosx=2/sin^2x

and
(1-cos^2x)(1+1/tan^2x)= 1

I haven't even gotten 'round to sny of the quedtions because the first one is just so hard. I'm not really sure I'm uderstanding how to use the quotient and pythagorean identities. I'm so confused. I can't make sense of it and I have tried so many different ways. I must have spent over an hour on he first problem and still I can't come up with an answer. I'm just really frustrated; could someone please help me. I'd really appreciate it.

  • Math - help really needed -

    i will do the first one

    1+1/tan^2x=1/sin^2x

    LS = 1 + 1/(sin^2x/cos^2x)
    = 1 + cos^2x/sin^2x
    = (sin^2x + cos^2x)/sin^2x
    = 1/sin^2x
    = RS

    I usually try to change all ratios to sines and cosines

  • Math - help really needed -

    the second is quite easy.

    add the left side terms by taking a common denominator of cosx

    you will get
    (1-cos^2x)/cosx
    = sin^2x/cosx
    = sinx(sinx/cosx)
    = sinxtanx
    = RS

  • Math - help really needed -

    PYTHAGOREAN IDENTITIES:

    sin^2 + cos^2 = 1 (a^2 + b^2 = c^2)

    sec^2 - tan^2 = 1
    csc^2 = cot^2 = 1 (the co-version of the above)


    (1/cosx)-(cosx/1)=sinxtanx

    To subtract those two, you need a common denominator, so multiply the (cosx/1) by (cosx/cosx). When you subtract them, the new numerator is an identity, so it can be re-written. When you simplify, you're left with sinxtanx.

    The rest are very similar. I hope you the idea of how to solve these types of problems - you just rearrange the complicated side to get identities that can be re-written.

  • Math - help really needed -

    oh, ok, I see.
    thank you all very much. I think I get it now, so hopefully I can actually get them on my own. :)

  • Math - help really needed -

    wait, for the first one if sin^2+cos^2=1, how did you even cancel some out?

  • Math - help really needed -

    I'll give you a hint:

    Try replacing the 1 with tan^2x/tan^2x. After add that to 1/tan^2x, replace the denominator (the tan^2x) with sin^2/cos^2.

  • Math - help really needed -

    okay, I'll try it out, thank you.

  • Math - help really needed -

    I'm really sorry. I don't want to sound stupid, but when I add tan^2x/tan^2x to 1/tan^2x I get 1+tan^2x/tan^2x and then the tans cancel out and I'm left with just one. Obviously, that's not correct, but I can't find my mistake.

  • Math - help really needed -

    You actually get (1/tan^2x)+1, which is cot^2 + (csc^2x-cot^2x), which is csc^2x, which is 1/sin^2x. And that's your answer.

    (You could only cancel out the tan^2 if it were (Some #)(tan^2x)/tan^2x, rather than (1)+(tan^2x)/tan^2x. Dividing is the inverse of multiplying, so dividing only cancels what is multiplied, not added.)

  • Math - help really needed -

    but if I have identical denominators, why can't I just add the 1 to the tan^2X?

  • Math - help really needed -

    1+tan^2x/tan^2x

    ^ I split that fraction, making the 1/denom. a fraction, and adding it to the tan^2/denom fraction.

  • Math - help really needed -

    I haven't learned what cot and csc are. Can I still solve this with just tan, cos, and sin?

  • Math - help really needed -

    cot= 1/tan or inverse of tangent
    csc= 1/sin or inverse of sine


    (1-cos^2x)(1+1/tan^2x)= 1

    (1-cos^2x)= sin x
    (1+1/tan^2x)= cscx

    So

    (sin x)(1/sin x)=1
    1=1

  • Math - help really needed -

    The first one is simpler if you do it this way:

    Multiply both sides by [sin(x)]^2
    then:
    1+1/tan^2x=1/sin^2x
    becomes

    sin^2x + sin^2x/tan^2x = 1

    since tan^2x = sin^2x/cos^2x the equation becomes:

    sin^2x + sin^2x/(sin^2x/cos^2x) = 1
    sin^2x + sin^2x*cos^2x/sin^2x = 1

    the two sin^2x cancels and you are left with:

    sin^2x + cos^2x = 1 which is true!

    Hope that helps!

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