tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
cot(115º) = -tan(25) or cot(115º) = -0.466308
First: note 3 things about cot and tan, and note the given statement:cot = 1/tantan is cyclic with a period of π, that is tan(nπ + x) = tan(x)tan is an odd function, that is tan(-x) = -tan(x)tan(π/4) = 1Now apply them to the problem:cot(π - π/4) = 1/tan(π - π/4)= 1/tan(-π/4)= 1/-tan(π/4)= 1/-1 = -1Thus:cot(π - π/4) = -1.
Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1
For an angle, ?, the cotangent (or cot) of the angle is given bycot ? = 1/tan ?If ?=65
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
whats the big doubt,cot/tan+1= 1+1= 2
Yes, it is.
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
The TI-83 does not have the cot button, however, if you type 1/tan( then this will work the same as the cot since cot=1/tan. The other way to do this is to type (cos(x))/(sin(x)) where x is the angle you're looking for. This works because cot=cos/sin
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
cot(115º) = -tan(25) or cot(115º) = -0.466308
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
cot(360°) = cot(0°) = tan(90°) = ∞
cot 115 deg = - tan25 deg
The reciprocal of the tangent is the cotangent, or cot. We might write 1/tan = cot.