solve: tan 15/2

 tan15/2 =sin15/2   / cos 15/2

cos 15/2  = sqrt[2+3^1/2]/2

sin15/2 =sqrt[2-3^1/2]/2

i.e tan 15/2  =sqrt[2-3^1/2]/sqrt[2+3^1/2]

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 Recall the following identities:

sin(x/2) = +/- Sqrt[(1-cos(x))/2]
cos(x/2) = +/- Sqrt[(1+cos(x))/2]
tan(x/2) = sin(x)/[1+cos(x)]

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I am assuming you want to prove the (true) statement: tan(7.5°) = √6 - √3 + √2 - 2; in order to do this, we need to use the half-angle formula for tangent: tan(θ/2) = sinθ/(1 + cosθ) = (1 - cosθ)/sinθ. To limit the number of denominator rationalizations needed, we should use: tan(θ/2) = (1 - cosθ)/sinθ. By letting θ = 15°, we get: tan(7.5°) = [1 - cos(15°)]/sin(15°). To find cos(15°) and sin(15°), we could use the sine and cosine half-angle formulas, but we have would radicals nested inside radicals, which would be hard to simplify; instead, we can use the sine and cosine subtraction formulas. We have: (a) cos(15°) = cos(45° - 30°), by writing 15° as a difference of two special angles = cos(45°)cos(30°) + sin(45°)sin(30°), by the cosine subtraction formula = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4 (b) sin(15°) = sin(45° - 30°), by doing the same as (a) = sin(45°)cos(30°) - cos(45°)sin(30°), by the sine subtraction formula = (√2/2)(√3/2) - (√2/2)(1/2) = (√6 - √2)/4. Thus: tan(7.5°) = [1 - cos(15°)]/sin(15°) = [1 - (√6 + √2)/4]/[(√6 - √2)/4], from above = (4 - √6 - √2)/(√6 - √2), by multiplying the numerator and denominator by 4 = [(4 - √6 - √2)(√6 + √2)]/(6 - 2), by rationalizing the denominator = (4√6 - 6 - 2√3 + 4√2 - 2√3 - 2)/4 = (4√6 - 4√3 + 4√2 - 8)/2 = √6 - √3 + √2 - 2, as required. I hope this helps!
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tan15/2 =sin15/2 / cos 15/2

cos 15/2 = sqrt[2+3^1/2]/2

sin15/2 =sqrt[2-3^1/2]/2

i.e tan 15/2 =sqrt[2-3^1/2]/sqrt[2+3^1/2]
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Tan 15 /2
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Tan15/2
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Tan 15/2
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