If cos2θ θ + 18 1 8 = sin2θ θ , find tanθ

Question 1 Report

If cos2 $θ$ + $\frac{1}{8}$ = sin2 $θ$ , find tan $θ$

\frac{3 \sqrt{7}}{7}

\sqrt{7}

\sqrt{7}

Answer Details

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

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Sine, Cosine And Tangent Of An Angle

Trigonometry

Modular Arithmetic

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