Can anybody explain tan(210deg - 45deg) and how it comes out to be sqrt3 - 2

Use tan(A - B) = [tanA - tanB] / [1 + tanA.tanB]


tan210° = 1/√3 and tan45° = 1

tan(210° - 45°) = [tan210° - tan45°] / [1 + tan210°.tan45°]

= [1/√3 - 1] / [1 + 1/√3]

= (1 - √3) / (1 + √3)

Multiply top and bottom by 1 - √3


= (1 - √3)(1 - √3) / (1 - √3)(1 + √3)

= (1 - 2√3 + 3) / (1 - 3)

= (4 - 2√3) / -2

= -(2 - √3)

= -2 + √3

= √3 - 2

Note:
tan(A - B) = (tanA - tanB)/(1 + tanAtanB)

tan(210 - 45)
= (tan210 - tan45)/(1 + tan210tan45)
= [(1/√3) - 1]/[1 + (1/√3)(1)]
= [(1/√3) - 1]/[(1/√3) + 1]
= [(1/√3)*√3/√3 - √3/√3]/[(1/√3)*√3/√3 + √3/√3]
= (√3/3 - √3/√3)/(√3/3 + √3/√3)
= (√3/3 - √3/√3*√3/√3)/(√3/3 + √3/√3*√3/√3)
= (√3 - 3)/3 / (√3 + 3)/3
= (√3 - 3)/3 * 3/(√3 + 3)
= (√3 - 3)/(√3 + 3)
= [(√3 - 3)/(√3 + 3)] * [(√3 - 3)/(√3 - 3)]
= [(√3 - 3)(√3 - 3)]/(3 - 9)
= [12 - 6√3] / -6
= [6(2 - √3)] / -6
= (2 - √3) / -1
= -1(2 - √3)
= √3 - 2

tan(210 - 45) = sin(210 - 45)/cos(210 - 45) =

=[sin 210cos45 - sin45 cos210)] / [cos 210 cos45 + sin 210 sin45] =

=[(-1/2 )* 1/√(2) - 1/√(2) (-√(3)/2)] / [(-√(3)/2) * 1/√(2) + (-1/2)*1/√(2)] =

= [ - 1/2√(2) +√(3)/2√(2)] / [ -√(3)/2√(2) - 1/2√(2)] =

= [ (√(3) - 1)/2√(2)] / [(-√(3) - 1)/2√(2)] =

=(√(3) - 1) / -(√(3 + 1) = multiply with the conjugate =

= - (√(3) - 1)² / (√(3+1) (√(3 - 1) = - [1 + 3 - 2√(3)] / 2 = √(3) - 2