waecmaths question:
If $\sin x=\tfrac{5}{13}$ and ${{0}^{\circ }}\le x\le {{90}^{\circ }}$ find the value of $(\cos x-\tan x)$
Option A:
$\tfrac{7}{13}$
Option B:
$\tfrac{12}{13}$
Option C:
$\tfrac{79}{156}$
Option D:
$\tfrac{209}{156}$
waecmaths solution:
$\begin{align} & \sin x=\tfrac{5}{13} \\ & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\ & \cos x=\sqrt{1-{{\sin }^{2}}x}=\sqrt{1-{{\sin }^{2}}x}=\sqrt{1-{{(\tfrac{5}{13})}^{2}}} \\ & \cos x=\sqrt{1-\tfrac{25}{169}}=\sqrt{\tfrac{144}{169}}=\tfrac{12}{13} \\ & \tan x=\frac{\sin x}{\cos x}=\frac{\tfrac{5}{13}}{\tfrac{12}{13}}=\frac{5}{12} \\ & \cos x-\tan x=\frac{12}{13}-\frac{5}{12}=\frac{144-65}{156}=\frac{79}{156} \\\end{align}$
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