If \(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\), find \(\tan \theta\).

Explanation

\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(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 \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)