Skin Care

Vietnam War

Animal Life

# How can remove a tan?

###### Wiki User

###### December 13, 2008 6:50PM

Stay out of the sun. It will fade with time. Don't try to use

chemicals; they damage your skin a lot more than it's worth.

## Related Questions

###### Asked in Skin Care, Beauty

### How do you remove fake tan?

###### Asked in Sun Tanning

### How do you remove suntan?

###### Asked in Visayan History

### How do you say mactan?

"How to say Mactan?", you asked. Well, here's how:
mək TäN' You say Mactan (as in the name of the Filipino Island)
like this:
Say Mac as in McDonald's --- the fast-food restaurant
Say tän as in Täng (it sounds like TOWN). By the way, Täng
is
A Chinese dynasty (618-907) that was known for its wealth and its
encouragement of the arts and literature.
Say Mactan at least 25 times, and you will remember it for
life:
mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN
mac TAN macTAN mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN mac
TAN mac TAN
mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN
mac TAN macTAN mac TAN mac TAN mac TAN mac TAN mac TAN mac TAN mac
TAN mac TAN

###### Asked in Trigonometry

### 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

###### Asked in Sun Tanning

### How do you remove tanning?

###### Asked in Math and Arithmetic, Algebra, Geometry

### 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).

###### Asked in Math and Arithmetic, Algebra, Trigonometry

### 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)