If 0^{circ}

Question

(B) Let $Z = \sin 	heta + \cos 	heta$.Squaring both sides,we get:$Z^{2} = (\sin 	heta + \cos 	heta)^{2} = \sin^{2} 	heta + \cos^{2} 	heta + 2 \s

Vedclass · Accepted Answer

Greater than $1$

Vedclass · Answer

(B) Let $Z = \sin heta + \cos heta$.Squaring both sides,we get:$Z^{2} = (\sin heta + \cos heta)^{2} = \sin^{2} heta + \cos^{2} heta + 2 \sin heta \cos heta$.Since $\sin^{2} heta + \cos^{2} heta = 1$,we have $Z^{2} = 1 + 2 \sin heta \cos heta$.For $0^{\circ} 0$.Therefore,$Z^{2} = 1 + ( ext{a positive value}) > 1$.Taking the square root,$Z > 1$.Thus,the value of $\sin heta + \cos heta$ is always greater than $1$ for $0^{\circ} < heta < 90^{\circ}$.

If $0^{\circ} < \theta < 90^{\circ},$ the value of $\sin \theta + \cos \theta$ is

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