✨ ベストアンサー ✨
0<5π/12<π/2より
sin(5π/12)>0かつcos(5π/12)>0なので
tan(5π/12)>0
よって
tan²(5π/12)={1/cos²(5π/12)}-1
tan(5π/12)=√({1/cos²(5π/12)}-1)
ここで半角の定理より
cos²(5π/12)
={cos(5π/6)+1}/2
=(1/2)(1-√3/2)
よって
1/cos²(5π/12)
={2/(1-√3/2)}×{(1+√3/2)/(1+√3/2)}
=2(1+√3/2)/(1-3/4)
=8(1+√3/2)
なので
{1/cos²(5π/12)}-1
=8(1+√3/2)-1
=7+4√3
=7+2√12
よって
tan(5π/12)
=√({1/cos²(5π/12)}-1)
=√(7+2√12)
=√4+√3
=2+√3