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

2\(\sqrt{2}\)

\(\sqrt{2}\)

\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{2}\)

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

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