If y+Cosθ=Sinθ has a real solution then

If $y + C o s θ = S i n θ$ has a real solution then

A
$- \sqrt{2} \leq y \leq \sqrt{2}$
B
$y > \sqrt{2}$
C
$y \leq - \sqrt{2}$
D
$y \leq 1$

Solution:

$y = \sin θ - \cos θ$ it has solution if $| c | \leq \sqrt{a^{2} + b^{2}} \Rightarrow - \sqrt{2} \leq y \leq \sqrt{2}$

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