Monique Slaughter

2021-12-31

If $\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}=4$, what is the value of $\mathrm{tan}\frac{x}{2}$?
1)-3
2)2
3)-2
4)3

Bertha Jordan

Let t be $t=\mathrm{tan}\left(\frac{x}{2}\right)$ for the "good x" satisfying the given relation. Then $\mathrm{sin}x=\frac{2t}{1+{t}^{2}}$, so
$4=\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}=\frac{\left(1+{t}^{2}\right)-2t}{\left(1+{t}^{2}\right)+2t}={\left(\frac{1-t}{1+t}\right)}^{2}$
This gives for $\frac{1-t}{1+t}$ the values $±2$, leading to the two solutions -3 and -1/3

maul124uk

You have
$\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}=4⇔\frac{{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)+{\mathrm{cos}}^{2}\left(\frac{x}{2}\right)-2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)}{{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)+{\mathrm{cos}}^{2}\left(\frac{x}{2}\right)+2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)}=4$
$⇔{\left(\frac{\mathrm{sin}\left(\frac{x}{2}\right)-\mathrm{cos}\left(\frac{x}{2}\right)}{\mathrm{sin}\left(\frac{x}{2}\right)+\mathrm{cos}\left(\frac{x}{2}\right)}\right)}^{2}=4$
$⇔\frac{\mathrm{sin}\left(\frac{x}{2}\right)-\mathrm{cos}\left(\frac{x}{2}\right)}{\mathrm{sin}\left(\frac{x}{2}\right)+\mathrm{cos}\left(\frac{x}{2}\right)}=±2$
$⇔\frac{\mathrm{tan}\left(\frac{x}{2}\right)-1}{\mathrm{tan}\left(\frac{x}{2}\right)+1}=±2$
The only solution of the equation $\frac{\mathrm{tan}\left(\frac{x}{2}\right)-1}{\mathrm{tan}\left(\frac{x}{2}\right)+1}=2$ is $\mathrm{tan}\left(\frac{x}{2}\right)=-3$ which is on that list, whereas the only solution of the equation $\frac{\mathrm{tan}\left(\frac{x}{2}\right)-1}{\mathrm{tan}\left(\frac{x}{2}\right)+1}=-2$ is $\mathrm{tan}\left(\frac{x}{2}\right)=-\frac{1}{3}$, which is not on that list.
So, the problem has two solutions, but only one of them is on the list of options.

karton

$\mathrm{tan}\frac{x}{2}=\frac{\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}}=\frac{2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}}{2{\mathrm{cos}}^{2}\frac{x}{2}}=\frac{\mathrm{sin}x}{\mathrm{cos}x+1}$
hence with

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