dai so

chung minh dang thuc
(1+Sin2x)/Cos2x=Tan(pi/4+x)

  1. 👍 1
  2. 👎 0
  3. 👁 3,747
  1. I assume you want to prove that it is an identity.

    LS = (sin^2 x + cos^2 x + 2sinxcosx)/(cos^2 x - sin^2 x)
    = (sinx + cosx)^2 / )(cosx+sinx)(cosx-sinx))
    = (sinx + cosx)/(cosx - sinx)

    RS = (tan π/4 + tanx)/( 1 - (tan π/4)(tanx) )
    = (1 + tanx)/(1 - tanx)
    = (1 + sinx/cosx) / 1 - sinx/cosx)
    multiply top and bottom by cosx

    = (cosx + sinx)/(cosx - sinx)
    = LS

    thus it is proven

    1. 👍 8
    2. 👎 7

Respond to this Question

First Name

Your Response

Similar Questions

  1. Trigonometry

    find sin2x, cos2x, and tan2x if sinx= -2/sqrt 5 and x terminates in quadrant III

  2. Trigonometry

    Simplify the expression using trig identities: 1. (sin4x - cos4x)/(sin2x -cos2x) 2. (sinx(cotx)+cosx)/(2cotx)

  3. Adv function

    Express sec2x in terms of tanx and secx I know you have to sec(2x) = 1/cos(2x) = 1/(cos²x - sin²x) But how do you split that. Like how to simplify that?

  4. Trigonometry

    Prove: 1-tanx/1+tanx=1-sin2x/cos2x

  1. Pre-calculus

    prove the identity sin2x-sin2y/sin2x+sin2y=tan(x-y)/tan(x+y)

  2. trigonometry

    if sin2x=3sin2y, prove that: 2tan(x-y)=tan(x+y) ( here, in sin2x, 2x is an angle., like there's a formula:sin2x=2sinxcosx and sin2y=2sinycosy; ....)

  3. PreCalculus

    Sin5x cos2x + cos5x sin2x=

  4. Mathematics

    1-sin2x/cosx=tan(π/4-x)

  1. math

    angle x lies in the third quadrant and tanx=7/24 determiner an exact value for cos2x determiner an exact value for sin2x

  2. trig

    tanx = 5/12 and sinx

  3. calc

    Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well, cos2x = cos2x - sin2x, so 1-coscx = 1 - cos2x - sin2x = 1 - cos2x + sin2x You should be able to

  4. Math

    Solve the equation 2cos2x = √3 for 0°≤x≤360° I did this: cos2x = √3 /2 2x=30 x=15 x=15, 165, 195, 345 Is this correct? Solve the equation √3 sin2x + cos2x = 0 for -π≤x≤π No idea how to approach this one Thanks a

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