In a ΔABC, b=3–√+1, c=3–√−1, ∠A=60°Δ𝐴𝐵𝐶, 𝑏=3+1, 𝑐=3-1, ∠𝐴=60°, then the value of tan(B−C2)tan𝐵-𝐶2 is

The given values are:

b = 3 + √1 = 4 c = 3 - √1 = 2 ∠A = 60°

We know that in any triangle, the sum of all angles is 180°. So, we can find ∠B and ∠C using the formula:

∠B + ∠C = 180° - ∠A ∠B + ∠C = 180° - 60° = 120°

We also know that in any triangle, the sides opposite to the larger angle is longer. So, ∠B > ∠C because b > c.

Now, we can use the sine rule which states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides of the triangle. So,

b/sinB = c/sinC 4/sinB = 2/sinC sinB/sinC = 2

We also know that tan(x) = sin(x)/cos(x). So,

tan(B - C) = sin(B - C)/cos(B - C) tan(B - C) = (sinB cosC - cosB sinC) / (cosB cosC + sinB sinC)

Substituting sinB/sinC = 2 into the equation, we get

tan(B - C) = (2 cosC - cosB) / (cosB + 2 cosC)

Since we don't have the values for cosB and cosC, we can't simplify further. But we can say that the value of tan(B - C) depends on the values of cosB and cosC.