atan
30°
Usually there is an inverse key or( tan -1 )key for this
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
The inverse of sine (sin) is cosecant (csc). The inverse of cosine (cos) is secant (sec). The inverse of tangent (tan) is cotangent (cot).
1
tan-1(0.5) = 26.6 degrees.
ArcTan is another name for ;Inverse Tan' or 'Tan^*-1) Hence ArcTan(0.55431) = 29.00004157 degrees. Effectively 29 degrees.
C. K. Tan has written: 'Strategic management'
There is not much that can be done by way of simplification. Suppose arccot(y) = tan(x) then y = cot[tan(x)] = 1/tan(tan(x)) Now cot is NOT the inverse of tan, but its reciprocal. So the expression in the first of above equation cannot be simplified further. Similarly tan[tan(x)] is NOT tan(x)*tan(x) = tan2(x)
d/dx[ tan-1(x) ] = 1/(1 + x2)
Two variables, x and y are in inverse variation if x*y = c for some constant c. The equation can be written in the form y = c/x.
-c+2d