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If cos2\(\theta\) + \(\frac{1}{8}\) = sin2\(\theta\), find tan\(\theta\)...


Question

If cos2\(\theta\) + \(\frac{1}{8}\) = sin2\(\theta\), find tan\(\theta\)

Options

A) 3

B) \(\frac{3\sqrt{7}}{7}\)

C) 3\(\sqrt{7}\)

D) \(\sqrt{7}\)

The correct answer is B.

Explanation:

cos2\(\theta\) + \(\frac{1}{8}\) = sin2\(\theta\)..........(i)
from trigometric ratios for an acute angle, where cos\(\theta\) + sin2\(\theta\) = 1 - cos\(\theta\) ........(ii)
Substitute for equation (i) in (i) = cos2\(\theta\) + \(\frac{1}{8}\) = 1 - cos2\(\theta\)
= cos2\(\theta\) + cos2\(\theta\) = 1 - \(\frac{1}{8}\)
2 cos2\(\theta\) = \(\frac{7}{8}\)
cos2\(\theta\) = \(\frac{7}{2 \times 3}\)
\(\frac{7}{16}\) = cos\(\theta\)
\(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\)
but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\)
opp2 = hyp2 - adj2
opp2 = 42 (\(\sqrt{7}\))2
= 16 - 7
opp = \(\sqrt{9}\) = 3
than \(\theta\) = \(\frac{\text{opp}}{\text{hyp}}\)
= \(\frac{3}{\sqrt{7}}\)
\(\frac{3}{\sqrt{7}}\) x \(\frac{7}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)

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Dicussion (1)

  • cos2\(\theta\) + \(\frac{1}{8}\) = sin2\(\theta\)..........(i)
    from trigometric ratios for an acute angle, where cos\(\theta\) + sin2\(\theta\) = 1 - cos\(\theta\) ........(ii)
    Substitute for equation (i) in (i) = cos2\(\theta\) + \(\frac{1}{8}\) = 1 - cos2\(\theta\)
    = cos2\(\theta\) + cos2\(\theta\) = 1 - \(\frac{1}{8}\)
    2 cos2\(\theta\) = \(\frac{7}{8}\)
    cos2\(\theta\) = \(\frac{7}{2 \times 3}\)
    \(\frac{7}{16}\) = cos\(\theta\)
    \(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\)
    but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\)
    opp2 = hyp2 - adj2
    opp2 = 42 (\(\sqrt{7}\))2
    = 16 - 7
    opp = \(\sqrt{9}\) = 3
    than \(\theta\) = \(\frac{\text{opp}}{\text{hyp}}\)
    = \(\frac{3}{\sqrt{7}}\)
    \(\frac{3}{\sqrt{7}}\) x \(\frac{7}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)

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