for asin θ + bcos θ + c,
maximum value = $$c+\sqrt{a^{2}+b^{2}}$$
minimum value = $$c-\sqrt{a^{2}+b^{2}}$$
for sin θ + cos θ , a = 1, b = 1, c = 0
maximum value = $$c+\sqrt{a^{2}+b^{2}}=0+\sqrt{1^{2}+1^{2}}=\sqrt{2}$$
so the answer is option B.
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