So, \( \sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x} \).
Then, \( (\sec x - \tan x)^2 = \left(\frac{1 - \sin x}{\cos x}\right)^2 = \frac{(1 - \sin x)^2}{\cos^2 x} \).
We also know that \( \cos^2 x = 1 - \sin^2 x \).
So,
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