1/tan(x+1)+1/tan(x-1)=1/tan8/31?

Question

solve

Answer 1

1/tan (x+1) + 1/tan (x-1) = 1/tan (8/31)

Expand tan (x+1) and tan (x-1) using the identity tan (x+y) = (tan x + tan y) / (1-tan x tan y):

(1-tan x tan 1)/(tan x + tan 1) + (1+tan x tan 1)/(tan x - tan 1) = 1/tan (8/31)

Multiply both sides by (tan x + tan 1) * (tan x - tan 1);

(1-tan x tan 1) * (tan x - tan 1) + (1+tan x tan 1) * (tan x + tan 1) = (tan x + tan 1) * (tan x - tan 1)/tan (8/31)

Simplify:

tan x - tan² x tan 1 - tan 1 + tan x tan² 1 + tan x + tan² x tan 1 + tan 1 + tan x tan² 1 = (tan² x - tan² 1) * cot (8/31)

2 tan x + 2 tan x tan² 1 = tan² x cot (8/31) - tan² 1 cot (8/31)

cot (8/31) tan² x - (2+2 tan² 1) tan x - tan² 1 cot (8/31) = 0

Applying the quadratic formula:

tan x = ((2+2 tan² 1) ± √((2+2 tan² 1)² + 4 cot² (8/31) tan² 1))/(2 cot (8/31)

x=arctan (((2+2 tan² 1) ± √((2+2 tan² 1)² + 4 cot² (8/31) tan² 1))/(2 cot (8/31)) ± πk
or
x=arctan (((2+2 tan² 1) ± √((2+2 tan² 1)² + 4 cot² (8/31) tan² 1))/(2 cot (8/31)) ± πk

x≈1.216693933567969 ± πk
x≈ -0.73097601210609 ± πk

Where k is an arbitrary integer.

Answer 2

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