Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)
Question 1 Report
Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)
Answer Details
Since tan x = 1, we can determine that x = 45 degrees or \(\frac{\pi}{4}\) radians.
Using the identity \(\tan^2x + 1 = \sec^2x\), we can determine that \(\sec x = \sqrt{2}\).
Next, we can use the identity \(\sin^2x + \cos^2x = 1\) to determine that \(\cos x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\).
Now we can substitute the values of \(\sec x\) and \(\cos x\) into the expression for \(\frac{1 - \sin^2 x}{\cos x}\) to get:
\[\frac{1 - \sin^2 x}{\cos x} = \frac{1 - \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}} = \boxed{\frac{\sqrt{2}}{2}}\]
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