A) 1
B) \[\frac{1}{2}\]
C) \[\frac{3}{2}\]
D) \[\frac{-1}{2}\]
Correct Answer: B
Solution :
[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.
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