If \(\cos x - \sin x = a\) and \(\cos x + \sin x = b\), then the value of \(\frac{1 - \tan^2 x}{1 + \tan^2 x}\) is
\(ab\)
Since \(1 + \tan^2 x = \sec^2 x\), the expression \(\frac{1 - \tan^2 x}{1 + \tan^2 x}\) can be rewritten as \(\frac{\cos^2 x - \sin^2 x}{1}= \cos^2 x - \sin^2 x\) after multiplying numerator and denominator by \(\cos^2 x\).
This can be factored as \(\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)\).
Substituting the given values: \((\cos x - \sin x)(\cos x + \sin x) = a \times b = ab\).
Hence, the value of the expression is \(ab\).
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