Tangent formula. The difference formula for tangent is: tan (x-y) = (tanx-tany)/ (1+t...

Tangent formula. The difference formula for tangent is: tan (x-y) = (tanx-tany)/ (1+tanxtany) So, tan (15^circ) = tan (45^circ-30^circ) = (tan (45^circ)-tan (30^circ))/ (1 eeeeehhhh eeeeehhhh 20,205 students helped 2 Answers 1 Asked eeeeehhhh. 986301369863014 years ago. The difference formula for tangent is: tan (x-y) = (tanx-tany)/ (1+tanxtany) So, tan (15^circ) = tan (45^circ-30^circ) = (tan (45^circ)-tan (30^circ))/ (1 Solution 1: By the formula A=rs, where A is the area of the triangle, r is the radius of the inscribed circle, and s is the semi-perimeter of the triangle, if we know the area and the semi-perimeter, we can find the radius. YC hasn't written a biography yet. f (x) = tan(x) Using the limit definition of the derivative, which states that the derivative of f (x) is given by f '(x) = lim Δx→0 f (x + Δx) − f (x) Δx, we have: f '(x) = lim Δx→0 tan(x +Δx) −tan(x) Δx Use the tangent addition formula tan(α +β) = tan(α) +tan(β) 1 − tan(α)tan(β): f '(x) = lim Δx→0 tan(x) +tan(Δx) 1−tan(x)tan(Δx) −tan(x) Δx Finding a common a = sqrt ( (-15+sqrt (865))/40) Compute f' (x) f' (x) = (2x)/ (x^2+1) The slope, m, of the tangent line at x = 2 is: m = f' (2) = (2 (4))/ ( (2)^2+1) m = 8/5 Compute g' (x) g' (x) = 4x^4+3x^2 The tangent lines are parallel when they have the same slope, therefore, we must set g' (x) equal to 8/5 8/5 = 4x^4+3x^2 20x^4+15x^2 -8 = 0 Let u = x^2 20u^2+15u -8 = 0 Use the quadratic formula but we See explanation. Feb 7, 2018 · (df)/(dx)=(x^2+4x)/(x+2)^2; equation of tangent is 3x-4y-2=0 and equation of normal is 4x+3y-11=0 To draw a tangent at a point (x_0,f(x_0)) on the curve f(x), we need to find the slope of the tangent, which is given by value of f'(x)=(df)/(dx) at that point i. Salman A. on Socratic peace YC gets smarter on Socratic YC joined Socratic 7. A science enthusiast! 1,622 students helped 1 Answer 1 Asked Amanda P. Feb 7, 2018 · (df)/(dx)=(x^2+4x)/(x+2)^2; equation of tangent is 3x-4y-2=0 and equation of normal is 4x+3y-11=0 To draw a tangent at a point (x_0,f(x_0)) on the curve f(x), we need to find the slope of the tangent, which is given by value of f'(x)=(df)/(dx) at that point i. lrhortfc wdnink haffnvn oeuc wozo vogtt smbgp emzhdv pmbn vwri