If in a triangle $ABC$, the altitudes from the vertices $A,B,C$ on opposite sides are in H.P., then $\sin A,\sin B,\sin C$ are in:

Let the altitudes be $h_a,h_b,h_c$.Area $\Delta =\frac{1}{2}a{h}_a=\frac{1}{2}b{h}_b=\frac{1}{2}c{h}_c$.Thus, $h_a=\frac{2\Delta }{a},h_b=\frac{2\Delta }{b},h_c=\frac{2\Delta }{c}$.Given $h_a,h_b,h_c$ are in H.P., so their reciprocals are in A.P.:$\frac{1}{h_a},\frac{1}{h_b},\frac{1}{h_c}$ are in A.P.$\frac{a}{2\Delta },\frac{b}{2\Delta },\frac{c}{2\Delta }$ are in A.P.This implies $a,b,c$ are in A.P.By Sine Rule: $a=2R\sin A,b=2R\sin B,c=2R\sin C$.Substituting these values:$2R\sin A,2R\sin B,2R\sin C$ are in A.P.$⟹\sin A,\sin B,\sin C$ are in A.P.