posted by .

We have to solve for the problem below to make sure that it is equivalent to 2csc(x). However, I keep getting 2/2sin(x) and not 2/sin(x), which is what 2csc(x) is.

tan(x)/(1+sec(x))+ (1+sec(x))/tan(x)= 2csc(x)

  • Math -

    tan(x)/(1+sec(x))+ (1+sec(x))/tan(x)
    = tan x/[(cosx+1)/cosx]
    + [(cosx+1)/cosx]/tanx
    = sin x/(cosx+1) + (cosx +1)/sinx
    = [sin^2 x + (cosx +1)^2]/[sinx(1 + cosx)]
    = (sin^2x + cos^2x + 2 cos x + 1)/[sinx(1 + cosx)]
    = 2 (1 + cosx)/[sinx(1 + cosx)]
    = 2/sin x = 2 csc x

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Integration

    Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct?
  2. more trig.... how fun!!!!

    if you can't help me with my first question hopw you can help me with this one. sec(-x)/csc(-x)=tan(x) thanx to anyone who can help From the definition of the sec and csc functions, and the tan function, sec(-x)/csc(-x) = sin(-x)/cos(-x) …
  3. math (trig)

    Prove: sin^2(x/2) = csc^2x - cot^2x / 2csc^2(x) + 2csc(x)cot(x) On the right, factor the numberator as a difference of two perfect squares. In the denominator, factor out 2cscx. You ought to prodeed rather quickly to the proof.
  4. Pre-cal/Trig

    (tan/(1+sec)) + ((1+sec)/tan) = 2csc (Show all work please.)
  5. Math Help

    1) 1+cos(3t)/ sin(3t) + sin(3t)/( 1+ cos(3t))= 2csc(3t) 2) sec^2 2u-1/ sec^2 2u= sin^2 2u 3) cosB/1- sinB= secB+ tanB
  6. trig

    For each expression in column I, choose the expression from column II to complete an identity: Column I Column II 1. -tanxcosx A. sin^2x/cos^2x 2. sec^2x-1 B. 1/sec^2x 3. sec x/cscx C. sin(-x) 4. 1+sin^2x^2x-cot^2x+sin^2x 5. …
  7. calculus (check my work please)

    Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)-1) dx ∫ tan(x)sec(x)[sec^4(x)-sec^2(x)] …
  8. calculus

    So I am suppose to evaulate this problem y=tan^4(2x) and I am confused. my friend did this : 3 tan ^4 (2x) d sec^ 2x (2x)= 6 tan ^4 (2x) d sec^2 (2x) She says it's right but what confuses me is she deriving the 4 and made it a three?
  9. Math

    Find an equation for the tangent line to the curve at (π/2 , 2). y = 4 + cot(x) - 2csc(x) I am confused how to take the derivative of this problem. When I tried to solve it I ended up with -csc^2 (x) + (2csc(x) * cot(x)). From …
  10. Pre-cal

    Solve (in radians) sec x = 2csc x When I work it out, it gets to 2cot x - 1 = 0 cot x = 1/2 Would it be ok to use tan^-1 (2) ?

More Similar Questions