0
atan
Wiki User
∙ 11y ago
Add your answer:
Earn +20 pts
What is the answer Tan inverse 0.5773?
30°
How do you find the inverse of a tangent on a calculator?
Usually there is an inverse key or( tan -1 )key for this
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
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).
What is reciprocal in trigonometry?
The inverse of sine (sin) is cosecant (csc). The inverse of cosine (cos) is secant (sec). The inverse of tangent (tan) is cotangent (cot).
What is 1 divided by c negative 1powered in algebra?
1
What is tan20tan32 plus tan32tan38 plus tan38tan20?
This may not be the most efficient method but ... Let the three angle be A, B and C. Then note that A + B + C = 20+32+38 = 90 so that C = 90-A+B. Therefore, sin(C) = sin[(90-(A+B) = cos(A+B) and cos(C) = cos[(90-(A+B) = sin(A+B). So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B) Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)] so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)] The given expressin is tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A) = tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B) substituting for cot(A+B) gives = tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)] cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression. = tan(A)*tan(B) + [1- tan(A)*tan(B)] = 1
What is the inverse of the tanent function 0.5?
tan-1(0.5) = 26.6 degrees.
What has the author C K Tan written?
C. K. Tan has written: 'Strategic management'
What is inverse variation?
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.
How can arccot of tanx be simplified?
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)
What is the derivative of inverse tangent of x?
d/dx[ tan-1(x) ] = 1/(1 + x2)
Write the additive inverse of?
-c+2d
