$$\sin 20^\circ \cos 70^\circ + \sin 70^\circ \cos 20^\circ$$ = $$\sin20^{\circ}\cos70^{\circ}+\cos20^{\circ}\sin70^{\circ}$$
This is in the form of $$\sin A\cos B+\cos A\sin B=\sin\left(A+B\right)$$
$$=$$> Â $$\sin20^{\circ}\cos70^{\circ}+\sin70^{\circ}\cos20^{\circ}=\sin\left(20^{\circ\ }+70^{\circ}\right)= \sin90^{\circ\ }=1$$
Hence, the correct answer is Option A
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