We have, rth^{ }term in the expansion of (xsinα+axcosα)10 is
Tr+1=Cr10(xsinα)10−r(xacosα)r
r=0,1,2,…,10
Tr+1 will be independent of x, when 10−2r=0
⇒r=5
Now, T6=C510(xsinα)5×(xacosα)5
=C510×a5×251(sin2α)5
T6 will be greatest, when sin2α=1
Therefore, C51025a5=C510
⇒a=2