William Boggs

2021-12-26

If $2\mathrm{sin}\theta +\mathrm{cos}\theta =\sqrt{3}$, what is the value of ${\mathrm{tan}}^{2}\theta +4\mathrm{tan}\theta$ ?

Melinda McCombs

${\left(2\mathrm{sin}\theta +\mathrm{cos}\theta \right)}^{2}=3$
$4{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta +4\mathrm{sin}\theta \mathrm{cos}\theta =3{\mathrm{sin}}^{2}\theta +3{\mathrm{cos}}^{2}\theta$
${\mathrm{sin}}^{2}\theta +4\mathrm{sin}\theta \mathrm{cos}\theta =2{\mathrm{cos}}^{2}\theta$
Now dividing both sides by ${\mathrm{cos}}^{2}\theta$
${\mathrm{tan}}^{2}\theta +4\mathrm{tan}\theta =2$

yotaniwc

Alternative: We have
$2\mathrm{sin}\theta =\sqrt{3}-\mathrm{cos}\theta$
Squaring,
$4-4{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta -2\sqrt{3}\mathrm{cos}\theta$
So,
$5{\mathrm{cos}}^{2}\theta -2\sqrt{3}\mathrm{cos}\theta -1=0$
Or
$\mathrm{cos}\theta =\frac{2\sqrt{3}±4\sqrt{2}}{10}$
Since further conditions haven't been given, both values would be valid, and from here $\mathrm{tan}\theta$ can be calculated.

karton

Simply square the constraint and you get $4{\mathrm{sin}}^{2}\left(x\right)+4\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=3$ then double angle and trig identity $3{\mathrm{sin}}^{2}\left(x\right)+2\mathrm{sin}\left(2x\right)=2$ furthermore this is equivalent to $2\mathrm{sin}\left(2x\right)=3{\mathrm{cos}}^{2}\left(x\right)-1$
and with your definition of the left hand side
${\mathrm{tan}}^{2}x+4\mathrm{tan}x=\frac{{\mathrm{sin}}^{2}x+2\mathrm{sin}2x}{{\mathrm{cos}}^{2}x}$
we can insert the definition of $2\mathrm{sin}\left(2x\right)$ we got from the constraint and we get
${\mathrm{tan}}^{2}x+4\mathrm{tan}x=\frac{{\mathrm{sin}}^{2}x+3{\mathrm{cos}}^{2}x-1}{{\mathrm{cos}}^{2}x}=\frac{2{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}=2$

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