Math - help really needed

posted by .

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

Prove each idenity.





(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


    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
    = sin^2x/cosx
    = sinx(sinx/cosx)
    = sinxtanx
    = RS

  • Math - help really needed -


    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)


    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 -


    ^ 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


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

  • Math - help really needed -

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

    Multiply both sides by [sin(x)]^2

    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!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Trig

    Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos …
  2. Trig.......

    I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr Reiny …
  3. Mathematics - Trigonometric Identities

    Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y …
  4. Math - help really needed

    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 …
  5. Math (trigonometric identities)

    I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx …
  6. Trig Identities

    Proving identities: 1) 1+ 1/tan^2x = 1/sin^2x 2) 2sin^2 x-1 = sin^2x - cos^2x 3) 1/cosx - cosx = sin x tan x 4) sin x + tan x =tan x (1+cos x) 5) 1/1-sin^2x= 1+tan^2 x How in the world do I prove this...please help... I appreciateyour …
  7. precalculus

    For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)-sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x)) c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y)) d. (tan(x)+sin(x))/(1+cos(x))=tan(x) …
  8. Trigonometry

    Prove the following trigonometric identities. please give a detailed answer because I don't understand this at all. a. sin(x)tan(x)=cos(x)/cot^2 (x) b. (1+tanx)^2=sec^2 (x)+2tan(x) c. 1/sin(x) + 1/cos(x) = (cosx+sinx)(secx)(cscx) d. …
  9. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know …
  10. Trig Identities

    Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1-sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) / (1-sin(x)) …

More Similar Questions