Math
posted by Shaun .
Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value.
(tan(π/2)+tan((2π)/3))
/
(1tan(π/2)tan((2π)/3))
Answer=__________
Exact value answer=_________
~Thank you.. I thought it was undefined, apparently not.
Respond to this Question
Similar Questions

Math
Evaluate *Note  We have to find the exact value of these. That I know to do. For example sin5π/12 will be broken into sin (π/6) + (π/4) So... sin 5π/12 sin (π/6) + (π/4) sin π/6 cos π/4 + cos … 
Math
Write an equation of the tangent function with period (3π)/8, phase shift π/5, and vertical shift 2. a. y=tan((8x)/3  (8π)/15) 2 b. y=tan((8x)/3 + (8π)/15) 2 c. y=tan((16x)/3 +(3π)/80) 2 d. y=tan((8x)/3 … 
Precalculus
Use one of the identities cos(t + 2πk) = cos t or sin(t + 2πk) = sin t to evaluate each expression. (Enter your answers in exact form.) (a) sin(19π/4) (b) sin(−19π/4) (c) cos(11π) (d) cos(53π/4) … 
Trigonometric
What is the maximum integer value of n, where n<120, that satisfies the following inequalities: sin(π/2+nπ/60)<0 and tan(π−nπ/60)<0? 
Calculus
How do I find the critical values? y= 4/x + tan(πx/8) What I did is I simplified it to y= 4x^1 + tan(πx/8) then I took the derivative y'= 4x^2 + (π/8)(sec(πx/8))^2 Then I simplied it y'= 4/x^2 + (π/8)(sec(πx/8))^2 
Math, please help
Which of the following are trigonometric identities? 
trigonometry
1) Find the exact value of the expression: tan−1(tan(−120651/47π))... How do you find Tan(120651/47pi)? 
precalc
Given that sec 3π/10 ≈ 17/10 and csc 3π/10 ≈ 17/14, find the following: 1. sin 3π/10 ≈ 14/17 (is this one correct) 2. csc 43π/10 ≈ 3. sec 2π/10 ≈ 4. cot 12π/10 ≈ 5. … 
calculus
1. what trig function has an amplitue of 1 and negative values for angles between π/2 and π? 
PreCal
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 …