Jessie Jenkins

2022-01-24

Reaching upon 9=1 while solving x for $3\mathrm{tan}\left(x-{15}^{\circ }\right)=\mathrm{tan}\left(x+{15}^{\circ }\right)$
Substituting $y=x+{45}^{\circ }$, we get
$3\mathrm{tan}\left(y-{60}^{\circ }\right)=\mathrm{tan}\left(y-{30}^{\circ }\right)$
$3\frac{\mathrm{tan}y-\sqrt{3}}{1+\sqrt{3}\mathrm{tan}y}=\frac{\mathrm{tan}y-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}\cdot \mathrm{tan}y}$

$3\left({\mathrm{tan}}^{2}y-3\right)=3{\mathrm{tan}}^{2}-1$

9=1

The solution provided by the book $x=n\pi +\frac{\pi }{4}$ fits, so why did i get 9=1?

### Answer & Explanation

You got
$0\cdot {\mathrm{tan}}^{2}y+9=1$
Id est, you got that $\mathrm{tan}y$ does not exist.
Thus,
$y=\frac{\pi }{2}+\pi n$
where $n\in \mathbb{Z}$, or
$x+{45}^{\circ }=\frac{\pi }{2}+\pi n$
which gives
$x=\pi n+\frac{\pi }{4}$

liep3p

Hint:
Set $\mathrm{tan}y=\frac{1}{a}$
$\frac{3\left(1-\sqrt{3}a\right)}{a+\sqrt{3}}=\frac{\sqrt{3}-a}{\sqrt{3}a+1}$
$⇔3-{a}^{2}=3\left(1-3{a}^{2}\right)⇔a=0$

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