Answer: [tex]\frac{sin (4x)}{cos(14 x)}[/tex]
Step-by-step explanation:
We have the following expression:
[tex]\frac{tan(9x)-tan(5x)}{1+tan(9x)tan(5x)}[/tex]
In order to work in a better way this expression, let's call [tex]\alpha=9x[/tex] and [tex]\beta=5x[/tex] and rewrite:
[tex]\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}[/tex]
Now, we know [tex]tan \theta=\frac{sin \theta}{cos \theta}[/tex]. Let's apply it here:
[tex]\frac{\frac{sin (\alpha)}{cos(\alpha)}-\frac{sin(\beta)}{cos(\beta)}}{1+\frac{sin (\alpha)}{cos(\alpha)}\frac{sin(\beta)}{cos(\beta)}}[/tex]
[tex]\frac{\frac{sin(\alpha)cos (\beta)-sin(\beta)cos(\alpha)}{cos(\alpha)cos(\beta)}}{\frac{cos(\alpha) cos(\beta)+sin(\alpha) sin(\beta)}{cos(\alpha)cos(\beta)}}[/tex]
Simplifying:
[tex]\frac{sin(\alpha)cos (\beta)-sin(\beta)cos(\alpha)}{cos(\alpha) cos(\beta)+sin(\alpha) sin(\beta)}[/tex]
According these trigonometric identities, we have:
[tex]sin(\alpha-\beta)=sin(\alpha)cos (\beta)-sin(\beta)cos(\alpha)[/tex]
[tex]cos(\alpha+\beta)=cos(\alpha) cos(\beta)+sin(\alpha) sin(\beta))[/tex]
Then:
[tex]\frac{sin(\alpha-\beta)}{cos(\alpha+\beta)}[/tex]
Writing again the actual values:
[tex]\frac{sin(9x-5x)}{cos(9x+5x)}[/tex]
Finally:
[tex]\frac{sin(4x)}{cos(14x)}[/tex]
