How do you use the sum and difference formula to simplify $ \sin 40\cos 10+\cos 40\sin 10 $ ?

Hint:
 We apply the theorem of the formula of sum and difference of angles for sin ratio. The formula is $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ . We put the values of $ A=40; B=10 $ in the equation to find the sum of the angles and the solution to the problem.

Complete step by step answer:
We know the formula of sum and difference of angles as
 $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ .
For our given problem we use the value of the angles as $ A=40; B=10 $ .
Placing the values, we get
 $ \sin 40\cos 10+\cos 40\sin 10=\sin \left( 40+10 \right)=\sin 50 $ .
We also know that $ \sin \alpha =\cos \left( 90-\alpha \right) $ . We put the value of $ \alpha $ as 50.
So, we can also write that $ \sin 50=\cos \left( 90-50 \right)=\cos 40 $ .
Therefore, the value of $ \sin 40\cos 10+\cos 40\sin 10 $ is $ \sin 50 $ .

Note:
We can also covert the sin ratios in cos ratios and make the equation in such a way that we can apply another formula of sum and difference of ratio cos to find the value of the problem. The formula is $ \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B $ .
We change $ \sin 40\cos 10+\cos 40\sin 10=\cos 50\cos 10+\sin 50\sin 10 $ , using the formula of $ \sin \alpha =\cos \left( 90-\alpha \right) $ .
We put the values of $ A=50;B=10 $
So, $ \cos 50\cos 10+\sin 50\sin 10=\cos \left( 50-10 \right)=\cos 40 $ . The solution is same.