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

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).

Tan Cerca...Tan Lejos was created in 1975.