If $$\sin \alpha + \cos \beta = 2;$$ $$\left( {{0^ \circ } \leqslant \beta Options: A . &nbsp$${\text{sin }}\frac{\alpha }{2}$$ B . &nbsp$$\cos \frac{\alpha }{3}$$ C . &nbsp$${\text{sin }}\frac{\alpha }{3}$$ D . &nbsp$$\cos \frac{{2\alpha }}{3}$$

If $$\sin \alpha + \cos \beta = 2;$$ $$\left( {{0^ \circ } \leqslant \beta <

Question

If $$\sin \alpha + \cos \beta = 2;$$ $$\left( {{0^ \circ } \leqslant \beta < \alpha \leqslant {{90}^ \circ }} \right){\text{,}}$$ then $${\text{sin}}\,{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }$$ is?

Options:

A . $${\text{sin }}\frac{\alpha }{2}$$

B . $$\cos \frac{\alpha }{3}$$

C . $${\text{sin }}\frac{\alpha }{3}$$

D . $$\cos \frac{{2\alpha }}{3}$$

Answer: Option B

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