• # question_answer 20) If $\alpha +\beta =\mathbf{9}{{\mathbf{0}}^{{}^\circ }}$, then the maximum value of sin $\alpha$.sin $\beta$. is A) 1                                             B) $\frac{1}{2}$              C) $\frac{3}{2}$                          D) $\frac{-1}{2}$

[b] $\alpha +\beta ={{90}^{{}^\circ }}$ $\because sin\alpha .sin\beta =\frac{1}{2}.2.sin\alpha .sin\beta$ $=\frac{1}{2}[cos\left( \alpha -\beta \right)-cos\left( \beta +\alpha \right)$ $=\frac{1}{2}\left[ cos\left( \alpha -\beta \right)-cos{{90}^{{}^\circ }} \right]$ $=\frac{1}{2}[cos\left( \alpha -\beta \right)$ $=\frac{1}{2}\times 1=\frac{1}{2}$ $\therefore$Maximum value of $cos\left( \alpha -\beta \right)=1$ Hence, option [b] is correct.