Yce Tan's birth name is Rycelonia Tan.
Chinkee Tan's birth name is Ferdinand Tan.
Teck Tan's birth name is Tan Chor Teck.
Guinness over Bass
Mika Tan :)
Of course! In fact, you should tan because it is said that having a tan makes you look better. Just make sure to use a light sunscreen to prevent from burning.
It is not good to tan too long but indoor tanning is more effective but outdoor is more safe.
Ok well ask sabrina1900 n she will not hack u but she will give ur the 66 outfit tan cheat n its well cool....
Hula Tan is the best suntan oil . Hula Tan is all natural and contains only the best oils for your body. I started Hula Tan last summer and big difference in my skin . No toxic chemicals just the good stuff . How do I know I am the one and only suntan girl .com . Yep Hula tan is my fav ok back to the beach !
Ya, I do it too but that is only cause I wear a bikini to tan every day.
you would get horrible tan lines
Hhahhahahaha you had a babyy??? bahahaha idn
Well...My mom works at a beauty salon, and people tan there, so I might know the anwser, not really sure, but I say No, because without the tanning bed, how you tan, well...you can tan at a beach...so it's a Yes, and a No in a way... I say No, if you disigree, it's OK, it's not like im will sue you...
probably not because unless ur skin is naturally tan, all that should happen after a hot bath is ur muscles being relaxed and if the water is very hot, burning fat off ur body. u can get tan obivously in the sun
tan(9) + tan(81) - tan(27) - tan(63) = 4
Tan Tan
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