$\begin{array}{l}
\dfrac{{\sin {{25}^o} + \cos {{70}^o}}}{{\sin {{20}^o} + \cos {{65}^o}}} = \dfrac{{\sin \left( {{{90}^o} - {{65}^o}} \right) + \cos \left( {{{90}^o} - {{20}^o}} \right)}}{{\sin {{20}^o} + \cos {{65}^o}}}\\
= \dfrac{{\cos {{65}^o} + \sin {{20}^o}}}{{\sin {{20}^o} + \cos {{65}^o}}} = 1
\end{array}$