waecmaths question:
Given that $\cos {{x}^{\circ }}=\tfrac{1}{r}$ express tan xo in terms of r
Option A:
$\tfrac{1}{\sqrt{r}}$
Option B:
$\sqrt{r}$
Option C:
$\sqrt{{{r}^{2}}+1}$
Option D:
$\sqrt{{{r}^{2}}-1}$
waecmaths solution:
$\begin{align} & \cos {{x}^{\circ }}=\frac{1}{r} \\ & Note:{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,\text{ } \\ & \sin x=\pm \sqrt{1-{{\cos }^{2}}x}=\pm \sqrt{1-{{(\tfrac{1}{r})}^{2}}} \\ & \text{We assume the angle is acute} \\ & \sin x=\sqrt{1-\tfrac{1}{{{r}^{2}}}}=\sqrt{\frac{{{r}^{2}}-1}{{{r}^{2}}}}=\frac{\sqrt{{{r}^{2}}-1}}{r} \\ & \tan x=\frac{\sin x}{\cos x}=\frac{\tfrac{\sqrt{{{r}^{2}}-1}}{r}}{\tfrac{1}{r}}=\frac{\sqrt{{{r}^{2}}-1}}{r}\times \frac{r}{1}=\sqrt{{{r}^{2}}-1} \\\end{align}$
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