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Sin theta 0 and tan theta 0?
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How do you solve tan squared theta - tan theta equals 0?
Let x = theta, since it's easier to type, and is essentially the same variable. Since tan^2(x)=tan(x), you know that tan(x) must either be 1 or zero for this statement to be true. So let tan(x)=0, and solve on your calculator by taking the inverse. Similarly for, tan(x)=1
What are the 6 trig functions for 0 degrees 90 degrees 180 degrees 270 degrees?
sin(0) = 0, sin(90) = 1, sin(180) = 0, sin (270) = -1 cos(0) = 1, cos(90) = 0, cos(180) = -1, cos (270) = 0 tan(0) = 0, tan (180) = 0. cosec(90) = 1, cosec(270) = -1 sec(0) = 1, sec(180) = -1 cot(90)= 0, cot(270) = 0 The rest of them: tan(90), tan (270) cosec(0), cosec(180) sec(90), sec(270) cot(0), cot(180) are not defined since they entail division by zero.
Why is tan90 is undefined?
tan90=Sin90/Cos90, sin90=1 and cos90=0 so 1/0 = undefined hence tan 90 is undifined
Which is greater tan 1tan2 tan3 .arrange them in descending order?
If the angles are measured in degrees or gradians, then: tan 3 > tan 2 > tan 1 If the angles are measured in radians, then: tan 1 > tan 3 > tan 2.
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).
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
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
How tall is Jaymee Tan?
Jaymee Tan is 6' 0".
What is the answer to cot squared x - tan squared x equals 0?
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.
How do you solve for the exact value of tan 2 pi?
tan 2 pi = tan 360º = 0
How do you make the left the same as the right for tan 0 times cot 0 all over tan 0 csc2 0?
You cannot because cot(0) is not defined, neither is cosec(0).
Sin theta 0 and tan theta 0?
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How is tan equal to undefined?
Tan of pi/2 + k*pi radians, for integer k, is not defined since tan = sin/cos and the cosine of these angles is 0. Since divsiion by 0 is not defined, the tan ratio is not defined.
How do you convert slope percent into tan degree value?
Let m be the slope in percent and theta be the angle in question. tan (theta) = m/100 theta = arctan (m/100) To verify the result, we know the following: tan 0 = 0 tan (45 degrees) = 1 = 100% tan (90 degrees) = infinity For example, if 0 < m < 100%, then 0 < theta < 45 degrees.
Why tan A 90 degree is not solve?
I am not sure what "tan A 90 degree" means. tan(90 degrees) is an expression that is not defined and so cannot be solved. One way to see why that may be so is to think of tan(x) = sin(x)/cos(x). When x = 90 degrees, sin(90) = 1 and cos(90)= 0 that tan(90) = 1/0 and since division by 0 is not defined, tan(90) is not defined.
What is the tangent of 45º?
0 is your answer tan(45)=1 and arccos(1)=0
Can sine theta equals tan theta equals theta be true for small angles?
If sin θ = tan θ, that means cos θ is 1 (since tan θ = (sin θ)/(cos θ)) (Usually in and equation a/b=a, b doesn't have to be 1 when a is 0, but cos θ = 1 if and only if sin θ = 0) The angles that satisfy cos θ = 1 is 2n(pi) (or 360n in degrees) When n is an integer. But if sin θ = tan θ = θ, the only answer is θ = 0. Because sin 0 is 0 and cos 0 is 1 and tan 0 is 0 The only answer would be when θ = 0.
