1. Write the general term

For $\left(x\sin\alpha + a\frac{\cos\alpha}{x}\right)^{10},$ the general term is $T_{r+1}=\binom{10}{r}(x\sin\alpha)^{10-r}\left(a\frac{\cos\alpha}{x}\right)^r.$

So, $T_{r+1}=\binom{10}{r}a^r(\sin\alpha)^{10-r}(\cos\alpha)^r x^{10-r-r}$ $=\binom{10}{r}a^r(\sin\alpha)^{10-r}(\cos\alpha)^r x^{10-2r}.$

2. Find the term independent of $x$

For the term independent of $x$, power of $x$ must be zero: $10-2r=0 \implies r=5.$

Thus the constant term is $T_6=\binom{10}{5}a^5(\sin\alpha)^5(\cos\alpha)^5.$

This can be written as $T_6=\binom{10}{5}a^5(\sin\alpha\cos\alpha)^5.$

3. Find its greatest value

We use $\sin\alpha\cos\alpha\le \frac12,$ with equality when $\alpha=45^\circ$.

Hence the greatest value of the constant term is $\binom{10}{5}a^5\left(\frac12\right)^5.$

Given that this greatest value equals $\frac{10!}{(5!)^2}=\binom{10}{5},$ we get $\binom{10}{5}a^5\left(\frac12\right)^5=\binom{10}{5}.$

Cancel $\binom{10}{5}$: $a^5\left(\frac12\right)^5=1$ $\left(\frac{a}{2}\right)^5=1.$

Therefore, $a=2.$

4. Check options

• A: $-1$ gives $\left(-\frac12\right)^5<0$, not possible for greatest value equal to positive number.
• B: $1$ gives $\left(\frac12\right)^5\ne 1$.
• C: $-2$ gives $(-1)^5=-1$, not possible.
• D: $2$ works.

So the correct option is D.