Trigo, caculus, maxima-minima,number theory- All in one

$\displaystyle n^2+4n\equiv 7({\rm mod}10)$ and so it implies $\displaystyle n^2+4n+3\equiv (n^2+4n-7)+(10)\equiv 10\equiv 0 ({\rm mod}10)$

Factorising we have, $\displaystyle (n+1)(n+3)\equiv 0 ({\rm mod}10)$ and since $n$ doesn't end with $7$ we have $n+1\equiv 0 ({\rm mod} 10) \implies n+3\equiv 2 ({\rm mod}10)$

And just a trickery to choose the option such that $\sin x+\cos y \le \boxed{2}$ which is our answer.

Trial and error works well enough here, once you recognize that n must be odd, since 4n is always going to be even. Try a few odd numbers:

3^2 + 4(3) = 21 5^2 + 4(5) = 45 9^2 + 4(9) = 117

Yep, that works! (Note that I didn’t try 7 because the question specifically says n can’t have a units digit of 7.) So n is 9, or 19, or 29, or whatever. In any case, the units digit of n + 3 is 2., which is the max value of sinx + cosy!!!