0
Tan(0) = 0
Remember
Tan(0) = Sin(0)/Cos(0) = 0/1
And Sin(0) = 0
Hence it follows that Tan(0) = 0
What else can I help you with?
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
Sin theta 0 and tan theta 0?
4
Is -tanA equals tanA?
The equation (-\tan A = \tan A) is true only when (\tan A = 0). This occurs at angles where (A) is an integer multiple of (\pi) (e.g., (0, \pi, 2\pi), etc.). In general, (-\tan A) is not equal to (\tan A) for most values of (A). Thus, the statement is not universally true.
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
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 the value of tan 0?
The value of tan 0 is 0. This is because the tangent function is defined as the ratio of the sine and cosine functions: tan(θ) = sin(θ)/cos(θ). At 0 degrees (or 0 radians), sin(0) is 0 and cos(0) is 1, resulting in tan(0) = 0/1 = 0.
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).
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.
Sin theta 0 and tan theta 0?
4
Is -tanA equals tanA?
The equation (-\tan A = \tan A) is true only when (\tan A = 0). This occurs at angles where (A) is an integer multiple of (\pi) (e.g., (0, \pi, 2\pi), etc.). In general, (-\tan A) is not equal to (\tan A) for most values of (A). Thus, the statement is not universally true.
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?
Tan(90) is infinitely large. Hence it is deemed not solved. Have a look at the graph of the Tan function. Youwill see it passes through '0' at an angle of 45 degrees. At 45 degrees , the graph line is steeper. At 89 degrees it is almost a vertical line. Hence a 90 degrees it is a vertical line and goes off the scale. Hence Tan(90) is unsresolved. Tan(0) = 0 Tan (45) = 1 Tan(60) = 1.7320... Tan(75) = 3.7320... Tan(85) = 11.430... Tan(89) = 57.289.... Tan(89.9) = 572.957... Tan(89.999....) = 57295779.51 Notice how the Tan values becomes larger and larger with increasing angle. Hence Tan(90) = unresolved; off the vertical scale.
