The value of

$\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$

$=\frac{\sin 70^{\circ}+\cos \left(90^{\circ}-50^{\circ}\right)}{\cos \left(90^{\circ}-20^{\circ}\right)+\sin 40^{\circ}}$  $\left[\because \cos \left(90^{\circ}-\theta\right)-\sin \theta\right]$

$=\frac{\sin 70^{\circ}+\sin 50^{\circ}}{\sin 20^{\circ}+\sin 40^{\circ}}$

$=\frac{2 \sin \left(\frac{70^{\circ}+50^{\circ}}{2}\right) \cos \left(\frac{70^{\circ}-50^{\circ}}{2}\right)}{2 \sin \left(\frac{20^{\circ}+40^{\circ}}{2}\right) \cos \left(\frac{40^{\circ}-20^{\circ}}{2}\right)} \quad\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

$=\frac{2 \sin 60^{\circ} \cos 10^{\circ}}{2 \sin 30^{\circ} \cos 10^{\circ}}$

$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$

$=\sqrt{3}$

Hence, $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}=\sqrt{3}$