Pre-Cal

posted by .

tan-1=43\9.3

  • Pre-Cal -

    totally unclear

    is there an argument after tan-1 ??

    did you mean tan^-1 Ø or something like that?

    is the right side = 43/9.3 ?

    I will assume you meant:

    tan^-1 Ø = 43/9.3
    = 4.6236...
    Ø = 77.796°

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Pre Cal.

    Evaluate tan(cos^(-1)(ã3/2 )+tan^(-1)(ã3/3))?
  2. Pre-Cal

    Perform the addition or subtraction and use the fundamental identities to simplify. tan x - sec^2 x / tan x I do not know where to start.
  3. Pre-Cal

    Perform the addition or subtraction. tanx - sec^2x/tanx tan^2(x)/tan(x) - sec^2x/tanx = tan^2x sec^2x / tanx then I use the identity 1+tan^2u=sec^2u I do not know what to do at this point.
  4. Pre-Cal

    1)tan 5 degrees + tan 25 degrees / 1 tan 5 degrees tan 25 degrees = sqrt 3 / 3 Am I correct?
  5. Pre-cal/Trig

    (tan/(1+sec)) + ((1+sec)/tan) = 2csc (Show all work please.)
  6. pre-cal

    tan((theta/2)+(pi/6))=-1
  7. Pre Cal

    cos^2+tan^2=1
  8. Pre Cal

    tan 2x- tan 6y/ 1+ tan 2x tan 6y
  9. Pre-Cal

    Use an Addition or Subtraction Formula to write the expression as a trigonometric function of one number. (tan(74°)-tan(14°))/(1+tan(74°)tan(14°)) =__________ Find it's exact value (not decimal form) =__________ Note: A similar …
  10. Pre-Cal

    Suppose that cosx= -3/5 and x is in the third quadrant, and that siny= 2/9 and y is in the second quadrant. Find the exact values 1). tan(y/2) 2). cos(x+y) 3). tan(x+y)

More Similar Questions