Question

I need help, I have tried looking it up but can’t find any help

Answer 1

Answer: $\frac{sin (4x)}{cos(14 x)}$

Step-by-step explanation:

We have the following expression:

$\frac{tan(9x)-tan(5x)}{1+tan(9x)tan(5x)}$

In order to work in a better way this expression, let's call $\alpha=9x$ and  $\beta=5x$ and rewrite:

$\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}$

Now, we know $tan \theta=\frac{sin \theta}{cos \theta}$. Let's apply it here:

$\frac{\frac{sin (\alpha)}{cos(\alpha)}-\frac{sin(\beta)}{cos(\beta)}}{1+\frac{sin (\alpha)}{cos(\alpha)}\frac{sin(\beta)}{cos(\beta)}}$

$\frac{\frac{sin(\alpha)cos (\beta)-sin(\beta)cos(\alpha)}{cos(\alpha)cos(\beta)}}{\frac{cos(\alpha) cos(\beta)+sin(\alpha) sin(\beta)}{cos(\alpha)cos(\beta)}}$

Simplifying:

$\frac{sin(\alpha)cos (\beta)-sin(\beta)cos(\alpha)}{cos(\alpha) cos(\beta)+sin(\alpha) sin(\beta)}$

According these trigonometric identities, we have:

$sin(\alpha-\beta)=sin(\alpha)cos (\beta)-sin(\beta)cos(\alpha)$

$cos(\alpha+\beta)=cos(\alpha) cos(\beta)+sin(\alpha) sin(\beta))$

Then:

$\frac{sin(\alpha-\beta)}{cos(\alpha+\beta)}$

Writing again the actual values:

$\frac{sin(9x-5x)}{cos(9x+5x)}$

Finally:

$\frac{sin(4x)}{cos(14x)}$