To simplify the given expression ( \frac{2(\sin 40^\circ + \cos 70^\circ)}{\cos 10^\circ} ), we first apply trigonometric identities to the numerator:

( \sin 40^\circ = \cos(90^\circ - 40^\circ) = \cos 50^\circ )

( \cos 70^\circ = \sin(90^\circ - 70^\circ) = \sin 20^\circ )

So, the expression becomes:

( \frac{2(\cos 50^\circ + \sin 20^\circ)}{\cos 10^\circ} )

Now, we use the sum-to-product trigonometric identity:

( \cos a + \sin b = \sqrt{2} \sin\left(\frac{\pi}{4} - \frac{a}{2}\right) )

Applying this identity, we have:

( \cos 50^\circ + \sin 20^\circ = \sqrt{2} \sin \left(\frac{\pi}{4} - \frac{50^\circ}{2}\right) + \sqrt{2} \sin \left(\frac{\pi}{4} - \frac{20^\circ}{2}\right) )

Simplifying further, we get:

( \sqrt{2} \sin \left(\frac{\pi}{4} - 25^\circ\right) + \sqrt{2} \sin \left(\frac{\pi}{4} - 10^\circ\right) )

Finally, evaluating the expression with the given trigonometric values and simplifying the denominator, we get:

( \frac{2\sqrt{2}(\sin 65^\circ + \sin 80^\circ)}{\cos 10^\circ} )