you could try going to a tanniing bed that keeps you tan for a lonmg time but i don't know about for the rest of your life? you could try going to a tanning bed that will keep you tan for a really long time but i don't know about the rest of your life?
no because all you are doing when you tan is darken the top layer of your epidermis so if you shave it wears off and in time it will wear off so no there is no permanent tan...
No, spray tanning is by far the safest way to get a tan. The dyes in spray tanning stain your skin for a short while and do no permanent damage.
If you mean a permanent tan, no. With enough sun exposure, you could certainly cause enough skin damage to generate a permanent discoloration, but it wouldn't be a uniform tan; it would much more likely be a cancerous spot or spots.
yes. it has happened to me and my recomendation is if you dont want an extremly dark tan is not to stand outside doing nothing for over 2 min. (just in case)
because a sun tan only effects the outer surface of the skin. a tattoo is much deeper down so the ink remains in the skin for a much longer period longer, given enoughf time a tattoo will fade as well.
tan(9) + tan(81) - tan(27) - tan(63) = 4
Tan Tan
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).
The airport code for Tan Tan Airport is TTA.
Tan Cerca...Tan Lejos was created in 1975.
cot(15)=1/tan(15) Let us find tan(15) tan(15)=tan(45-30) tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b)) tan(45-30)= (tan(45)-tan(30))/(1+tan(45)tan(30)) substitute tan(45)=1 and tan(30)=1/√3 into the equation. tan(45-30) = (1- 1/√3) / (1+1/√3) =(√3-1)/(√3+1) The exact value of cot(15) is the reciprocal of the above which is: (√3+1) /(√3-1)
tan(135) = -tan(180-135) = -tan(45) = -1