Answer:
The simplest form is tan(4x)
Step-by-step explanation:
* Lets revise the identity of the compound angles
- [tex]tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)tan(b)}[/tex]
- [tex]tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}[/tex]
* Lets solve the problem
- Let 9x = 5x + 4x
∴ tan(9x) = tan(5x + 4x)
- Use the rule of the compound angle
∵ [tex]\frac{tan(9x)-tan(5x)}{1+tan(9x)tan(5x)}[/tex] ⇒ (1)
∵ [tex]tan(5x+4x)=\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)}[/tex] ⇒ (2)
∵ tan(9x) = equation (2)
- Substitute (2) in (1)
∴ [tex]\frac{\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)}-tan(5x)}{1+(\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)})tan(5x)}[/tex]
- Multiply up and down by (1 - tan(5x)tan(4x))
∴ [tex]\frac{tan(5x)+tan(4x)-tan(5x)[1-tan(5x)tan(4x)]}{1-tan(5x)tan(4x)+tan(5x)[tan(5x)+tan(4x)]}[/tex]
- Simplify up and down
∴ [tex]\frac{tan(5x)+tan(4x)-tan(5x)+tan^{2}(5x)tan(4x)}{1-tan(5x)tan(4x)+tan^{2}(5x)+tan(5x)tan(4x) }[/tex]
∴ [tex]\frac{tan(4x)+tan^{2}(5x)tan(4x)}{[1+tan^{2}(5x)]}[/tex]
- Take tan(4x) as a common factor up
∴ [tex]\frac{tan(4x)[1+tan^{2}(5x)]}{[1+tan^{2}(5x)]}[/tex]
- Cancel [1 + tan²(5x)] up and down
∴ The answer is tan(4x)
