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Given that tan x = 1, where 0^o $\geq$ x 90^...

Given that tan x = 1, where 0^o $\geq$ x 90^o, evaluate $\frac{1 - \sin^2 x}{\cos x}$

A.

2 $\sqrt{2}$

B.

$\sqrt{2}$

C.

$\frac{\sqrt{2}}{2}$

D.

$\frac{1}{2}$

View answer & explanation

Correct answer is C

Given tan x = 1

x = tan^-1(1)

x = 45^o

Now, $\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}$

= $\frac{1 - \frac{1}{2}}{\frac{1}{2}}$

= $\frac{1}{2} + \frac{1}{\sqrt{2}}$

= $\frac{1}{2} \times \frac{1}{\sqrt{2}}$

= $\frac{\sqrt{2}}{2}$

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