Solve the equation $\frac{1}{(\mathrm{cos}\theta {)}^{2}}=2\sqrt{3}\mathrm{tan}\theta -2$

Damion Hardin
2022-05-09
Answered

Solve the equation $\frac{1}{(\mathrm{cos}\theta {)}^{2}}=2\sqrt{3}\mathrm{tan}\theta -2$

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rynosluv101swv2s

Answered 2022-05-10
Author has **19** answers

Your equation can be written as

$\frac{1}{{\mathrm{cos}}^{2}(x)}=$

$1+{\mathrm{tan}}^{2}(x)=2\sqrt{3}\mathrm{tan}(x)-2$

or

${\mathrm{tan}}^{2}(x)-2\sqrt{3}\mathrm{tan}(x)+3=0$

the reduced discriminant is

$\delta =3-3=0$

thus, there is one solution given by

$\mathrm{tan}(x)=\sqrt{3}$ which gives

$x=\frac{\pi}{3}.$

$\frac{1}{{\mathrm{cos}}^{2}(x)}=$

$1+{\mathrm{tan}}^{2}(x)=2\sqrt{3}\mathrm{tan}(x)-2$

or

${\mathrm{tan}}^{2}(x)-2\sqrt{3}\mathrm{tan}(x)+3=0$

the reduced discriminant is

$\delta =3-3=0$

thus, there is one solution given by

$\mathrm{tan}(x)=\sqrt{3}$ which gives

$x=\frac{\pi}{3}.$

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Why is $\mathrm{tan}\theta \approx \frac{1}{\frac{\pi}{2}-\theta}$ for $\theta$ close to $\frac{\pi}{2}$ ?

I wanted to see what the behaviour of the steep part of the$\mathrm{tan}$ curve was like, i.e. the behaviour of $\mathrm{tan}\left(x\right)\text{}as\text{}x\to {\left(\frac{\pi}{2}\right)}^{-}$ . So by thinking about a shift of the graph of $\mathrm{tan}\left(x\right)\text{}by\frac{\pi}{2}$ to the left, I put some small (positive and negative) values of $\theta$ into my calculator for the function $\mathrm{tan}(\theta +\frac{\pi}{2})$

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But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of$\mathrm{tan}\left(x\right)$ . Is there a more intuitive explanation? I couldn't think of any explanations analogous to those explaining small angle approximations.

I wanted to see what the behaviour of the steep part of the

or, in more colloquial terms,

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