math

prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx

  1. 👍 4
  2. 👎 0
  3. 👁 1,078
  1. LHS = (2sinxcosx-cosx)/(1-sinx-cos^2x+sin^2x)
    = (2sinxcosx-cosx)/(sin^2x+cos^2x-sinx-cos^2x+sin^2x)
    = (2sinxcosx-cosx)/(2sin^2x+cos^2x)
    = cosx(2sinx-1)/sinx(2sinx-1)
    = cosx/sinx
    = cotx
    =RHS

    1. 👍 2
    2. 👎 6

Respond to this Question

First Name

Your Response

Similar Questions

  1. Pre-calculus

    Prove the following identities. 1. 1+cosx/1-cosx = secx + 1/secx -1 2. (tanx + cotx)^2=sec^2x csc^2x 3. cos(x+y) cos(x-y)= cos^2x - sin^2y

  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

  3. math;)

    The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the

  4. Math

    How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX

  1. Math - Trig - Double Angles

    Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My

  2. 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)) /

  3. Math help again

    cos(3π/4+x) + sin (3π/4 -x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx = -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx) I canceled out -1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx +

  4. Trigonometry

    Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do.

  1. Math, please help

    Which of the following are trigonometric identities? (Can be more then one answer) tanx cosx cscx = 1 secx-cosx/secs=sin^2x 1-tanxtany=cos(x+y)/cosxcosy 4cosx sinx = 2cosx + 1 - 2sinx Find all solutions to the equation cosx

  2. trig

    express this in sinx (1/ cscx + cotx )+ (1/cscx- cotx) i got 2sinx is that right?? and B) express in cosx problem: is 1 + cotx/cscx - sin^2x i get to the step of 1 + cos-sin^2x and im stuck..help! (1/cscx + cotx) + (1/cscx - cotx)

  3. trig

    (cotxsec2x - cotx)/ (sinxtanx + cosx) = sinx prove. i keep getting (if i start from the left side) = cosx instead of sinx please help!

  4. Math

    1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) tan(4(3.14)/3) d) arccos(- square root of three/2) e) csctheata=2 2) verify the following identities: a) cotxcosx+sinx=cscx b)[(1+sinx)/ cosx] + [cosx/

You can view more similar questions or ask a new question.