[tex]\large\underline{\sf{Solution-}}[/tex]

Consider,

[tex]\sf \: \dfrac{sin4x + sin5x + sin6x}{cos4x + cos5x + cos6x} \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \: \dfrac{(sin6x + sin4x) + sin5x}{(cos6x + cos4x) + cos5x} \\ \\ [/tex]

We know,

[tex]\boxed{\begin{aligned}& \qquad \:\sf \:sinx + siny = 2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \qquad \: \\ \\& \qquad \:\sf \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \end{aligned}} \qquad \: \\ \\ [/tex]

So, using these results, the above expression can be rewritten as

[tex]\sf \: = \: \dfrac{2sin\bigg(\dfrac{6x + 4x}{2} \bigg)cos\bigg(\dfrac{6x - 4x}{2} \bigg) + sin5x}{2cos\bigg(\dfrac{6x + 4x}{2} \bigg)cos\bigg(\dfrac{6x - 4x}{2} \bigg) + cos5x} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{2sin5xcosx + sin5x}{2cos5xcosx + cos5x} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{sin5x(2cosx + 1)}{cos5x(2cosx + 1)} \\ \\ [/tex]

[tex]\sf \: = \: tan5x \\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: \dfrac{sin4x + sin5x + sin6x}{cos4x + cos5x + cos6x} = tan5x \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\boxed{\begin{aligned}& \qquad \:\sf \:sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \qquad \:\\ \\& \qquad \:\sf \: sinx  -  siny=2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg) \\ \\& \qquad \:\sf \: cosx + cosy=2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\\ \\& \qquad \:\sf \: cosx -  cosy=2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{y - x}{2} \bigg) \end{aligned}} \qquad  \\  \\[/tex]