\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i)
from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii)
Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \)
= \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\)
\(2\cos^2 \theta = \frac{7}{8}\)
\(\cos^2 \theta = \frac{7}{2 \times 8}\)
\(\frac{7}{16} = \cos \theta\)
\(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\)
but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\)
\(opp^2 = hyp^2 - adj^2\)
\(opp^2 = 4^2 - (\sqrt{7})^{2}\)
= 16 - 7
opp = \(\sqrt{9}\) = 3
\(\tan \theta = \frac{\text{opp}}{\text{adj}}\)
= \(\frac{3}{\sqrt{7}}\)
\(\frac{3}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)
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