Question 3

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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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