If 0° < θ < 90° ; the value of sin θ + cos θ is?

greater than 1

sin θ + cos θ = \(\sqrt {2 }\) (\(\frac{1}{\sqrt {2 }}\) sin θ + \(\frac{1}{\sqrt {2 }}\)cos θ

so multiplying and dividing with \(\sqrt {2 }\)

                   = \(\sqrt {2 }\) (sin θ cos 45° + cos θ sin 45°)

                   = \(\sqrt {2 }\) (sin (θ + 45°))

since 0° < θ < 90° ⇒ 45° < θ + 45° < 90° + 45° at minimum value θ ; θ = 0°

value of function sin(θ + 45°) = sin 45° = \(\frac{1}{\sqrt {2 }}\)

∴ minimum value is \(\sqrt {2}\) × \(\frac{1}{\sqrt {2 }}\) = 1

∴ value of sin θ + cos θ is always greater than 1