Show that $\displaystyle a^{\phi(b)}+b^{\phi(a)} \equiv 1 \pmod{ab}$ if a and b are relatively prime.

Starting with a and b are relatively prime I get

$\displaystyle a^{\phi(b)} \equiv 1 \pmod{b}

$

$\displaystyle

b^{\phi(a)} \equiv 1 \pmod{a}

$

But I am not sure how to combine the moduli to get $\displaystyle \pmod{ab}$. Would Chinese Remainder Theorem be the right approach?