waecmaths question:
Given that $\tan x=1$ where ${{0}^{\circ }}\le x\le {{90}^{\circ }}$ evaluate $\frac{1-{{\sin }^{2}}x}{\cos x}$
Option A:
$2\sqrt{2}$
Option B:
$\sqrt{2}$
Option C:
$\frac{\sqrt{2}}{2}$
Option D:
$\frac{1}{2}$
waecmaths solution:
$\begin{align} & \tan x=\frac{\cos x}{\sin x}=\frac{1}{1} \\ & \cos x=1,\text{ }\sin x=1 \\ & Note:co{{s}^{2}}x=1-{{\sin }^{2}}x \\ & \frac{1-{{\sin }^{2}}x}{\cos x}=\frac{{{\cos }^{2}}x}{\cos x}=\cos x=1 \\\end{align}$
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