$sin x + cos x = 1 + sin x . cos x$ , if

sin (x + \frac{π}{4}) = \frac{1}{\sqrt{2}}

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sin (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}

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cos (x + \frac{π}{4}) = \frac{1}{\sqrt{2}}

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cos (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}

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Solution

The correct options are
A $sin (x + \frac{π}{4}) = \frac{1}{\sqrt{2}}$
B $cos (x + \frac{π}{4}) = \frac{1}{\sqrt{2}}$
D $cos (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}$
$sin (x) + cos (x) - 1 - sin (x) cos (x) = 0$
$sin (x) [1 - cos (x)] - (1 - cos (x)) = 0$
$(1 - cos (x)) (sin (x) - 1) = 0$
Hence
$cos (x) = 1$ or $x = 2 k π$ and
$sin (x) = 1$ or $x = 2 k π + \frac{π}{2}$ .
For $k = 0, x = 0 o r π / 2 \Rightarrow$ options A,C,D are true

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