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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?
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∙ 14y ago
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Is it possible to get a perminant tan darker?
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...
Does a spray tan harm your skin?
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.
Does the sun permanently darken skin if exposure is during childhood?
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.
In animal crossing wild world is your tan permanent?
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)
Why is a tattoo permanent and a sun tan not?
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 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
What is a antonym of pale?
Tan Tan
What is tan20tan32 plus tan32tan38 plus tan38tan20?
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
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).
When was Tan Cerca...Tan Lejos created?
Tan Cerca...Tan Lejos was created in 1975.
What is the airport code for Tan Tan Airport?
The airport code for Tan Tan Airport is TTA.
What is the exact trigonometric function value of cot 15 degrees?
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)
