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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so the answer is option B.