You can use the double angle formula
[tex]\sin 2x = 2\sin x \cos x[/tex] and [tex]\cos 2x = 1 - 2\sin^2 x[/tex]
and the angle shift identity:
[tex]\cos(90-x) = \sin x\\\sin (90-x) = \cos x[/tex]
So:
[tex]\sin 10 + \frac{\sin 40}{\cos 40} \cos 10 = \\\sin 10 + \frac{\sin 40}{\cos 40} \sin 80 =\\ \sin 10 + \frac{\sin 40}{\cos 40} 2 \sin 40 \cos 40 = \\\sin 10 + 2 \sin ^2 40 = \\\cos 80 + \frac{2(1-\cos 80)}{2} = 1\\[/tex]
