trig

prove that (sin^4x-cos^4x)/(sinx-cosx) = sinx+cosx

  1. 👍 0
  2. 👎 0
  3. 👁 121
  1. (sin^4x-cos^4x)/(sinx-cosx)
    =(sin^2x+cos^2x)(sin^2x-cos^2x)/(sinx-cosx)
    =1*(sinx+cosx)(sinx-cosx)/(sinx-cosx)
    =sinx+cosx

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. math

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

  2. 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

  3. 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

  4. TRIGONOMETRY *(MATHS)

    Q.1 Prove the following identities:- (i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx (ii) (1+cotx+tanx)(sinx-cosx)/sec^3x-cosec^3x = sin^2xcos^2x.

  1. maths

    (Sin^3x-cos^3x)/(sinx-cosx) – cosx/sqrt(1+cot^2x)-2tanxcotx=-1 where x∈(0,2pi) general value of x.

  2. 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 +

  3. 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

  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/

  1. trigonometry

    how do i simplify (secx - cosx) / sinx? i tried splitting the numerator up so that i had (secx / sinx) - (cosx / sinx) and then i changed sec x to 1/ cosx so that i had ((1/cosx)/ sinx) - (cos x / sinx) after that i get stuck

  2. 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

  3. Maths

    1. If x is an acute angle and tan x =3/4, evaluate (a) cosx - sinx (b) cosx + sinx cosx+sinx cosx - sinx

  4. precal

    1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx) =sinx/cosx+ sinx /cosx= -2tanxI but I know this can't be correct because what I did doesn't end as a

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