How do you simplify \[\dfrac{{\sin 6x}}{{\sin 5x}}\]?

Hint: We will use the fact that 6x = x + 5x and then use the formula for Sin(a + b) in the numerator of the given expression and thus keep on simplifying to get the required answer.

Complete step-by-step answer:
We are given that we are required to simplify \[\dfrac{{\sin 6x}}{{\sin 5x}}\].
We know that 6x = 5x + x.
Now taking sine function of both the sides, we will then get the following expression:-
$ \Rightarrow $sin (6x) = sin (5x + x)
We will now use the formula given by the following equation:-
$ \Rightarrow $sin (a + b) = sin a . cos b + cos a . sin b
Replacing a by 5x and b by x, we will then get the following expression:-
$ \Rightarrow $sin (5x + x) = sin 5x . cos x + cos 5x . sin x
Putting this in the given expression \[\dfrac{{\sin 6x}}{{\sin 5x}}\], we will then obtain:-
\[ \Rightarrow \dfrac{{\sin 6x}}{{\sin 5x}} = \dfrac{{\sin 5x\cos x + \cos 5x\sin x}}{{\sin 5x}}\]
Now, we will use the fact that: $\dfrac{{a + b}}{c} = \dfrac{a}{c} + \dfrac{b}{c}$.
\[ \Rightarrow \dfrac{{\sin 6x}}{{\sin 5x}} = \dfrac{{\sin 5x\cos x + \cos 5x\sin x}}{{\sin 5x}} = \dfrac{{\sin 5x\cos x}}{{\sin 5x}} + \dfrac{{\cos 5x\sin x}}{{\sin 5x}}\]
Cancelling out the sin 5x in the first term of addition in the right hand side, we will then get:-
\[ \Rightarrow \dfrac{{\sin 6x}}{{\sin 5x}} = \cos x + \sin x\dfrac{{\cos 5x}}{{\sin 5x}}\]
Now, we will use the fact that: $\dfrac{{\cos x}}{{\sin x}} = \cot x$
\[ \Rightarrow \dfrac{{\sin 6x}}{{\sin 5x}} = \cos x + \sin x\cot 5x\]

Hence, we have the simplification as cos x + sin x . cot 5x

Note:
The students must note that if they broke off both numerator and denominator in two parts as we did in denominator, then we would not be able to separate it out as we did and cancelled out. Therefore, we broke off the one which was larger as we could break 6x in such a way that we cancelled 5x or something. But, we cannot break 5x such that we induce a 6x in between.
The students must commit to memory the following formulas:-
$ \Rightarrow $sin (a + b) = sin a . cos b + cos a . sin b
$ \Rightarrow \dfrac{{\cos x}}{{\sin x}} = \cot x$
$ \Rightarrow \dfrac{{a + b}}{c} = \dfrac{a}{c} + \dfrac{b}{c}$