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How do you solve for the exact value of tan 2 pi?
tan 2 pi = tan 360º = 0
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).
Find the value of tan 180 plus 15?
For tan(180 degrees), this is simply sin(180 degrees)/cos(180 degrees). To find these values, note that 180 degrees is the leftmost point on the unit circle, at y=0, x=-1, so is tan(180 degrees)=0/-1=0. Then adding 15 gives 15.
What is the value of tan 48degrees19 23?
= tan (48.323 deg) = 1.1232
What is the value of tan 30?
tan (30 degrees) would be equal to 0.577350269.
How do you solve for the exact value of tan 2 pi?
tan 2 pi = tan 360º = 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 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.
What is tan of 0?
Tan of 0 equals zero.
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 -90?
Assuming you mean -90 degrees, not radians: tan (-90) = [sin(-90)]/[cos(-90)] = (-1) / 0 You cannot divide by zero. tan (-90) is undefined/does not exist.
What is the exact trigonometric function value of cot 0?
There is no value cot 0, because cot 0 is equivalent to 1 / tan 0, which is equivalent to 1 / 0, which is undefined. That said, the limit of cot x as x approaches 0 is infinity.
By using trigonometric identities find the value of sin A if tan A a half?
The value of tan A is not clear from the question.However, sin A = sqrt[tan^2 A /(tan^2 A + 1)]
How do you find the value of tan 135?
tan(135) = -tan(180-135) = -tan(45) = -1
Find the value of tan 180 plus 15?
For tan(180 degrees), this is simply sin(180 degrees)/cos(180 degrees). To find these values, note that 180 degrees is the leftmost point on the unit circle, at y=0, x=-1, so is tan(180 degrees)=0/-1=0. Then adding 15 gives 15.
How tall is Jaymee Tan?
Jaymee Tan is 6' 0".
What is the value of tan 22.5 degrees?
tan(22.5)=0.414213562
