Question 25

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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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