Help me solve this

[hard nebula]

Well, did you do what's written?

[fossil lichen]

note that

$\mathrm{cos}^2(\theta} = \frac{1}{\mathrm{tan}^2(\theta) + 1}$

[steady sorrel]

borekking
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[fossil lichen]

$\mathrm{cos}^2(\theta) = \frac{1}{\mathrm{tan}^2(\theta) + 1}$