Maths

If is a n acute angle and tanx=3 4 evaluate cosx-sinx cosx+sinx

  1. 👍 1
  2. 👎 0
  3. 👁 882
  1. if tanx = 3/4,
    sinx = 3/5
    cosx = 4/5

    and go from there

    1. 👍 3
    2. 👎 4
  2. the ans

    1. 👍 2
    2. 👎 0
  3. 2+2-4

    1. 👍 0
    2. 👎 1
  4. 4/5-3/5 divided by 4/5+3/5
    Then go from there

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Further maths

    If x is an acute angle, tanx =3/4,evaluate coax _sinx÷cosx+sinx

  2. 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

  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. 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

  1. Trig

    Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx cosx - sinx/cosx =(2sinxcos^2 x - sinx)/cosx =sinx(2cos^2 x -1)/cosx = L.S. Q.E.D.

  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. 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)) /

  4. 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

  1. Math

    (sinx - cosx)(sinx + cosx) = 2sin^2x -1 I need some tips on trigonometric identities. Why shouldn't I just turn (sinx + cosx) into 1 and would it still have the same identity?

  2. 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.

  3. Calculus

    determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0

  4. Maths

    Solve this equation fo rx in the interval 0

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