A) \[sin\,{{15}^{{}^\circ }}\]
B) \[cos\,{{15}^{{}^\circ }}\]
C) \[sin\text{ }{{15}^{{}^\circ }}\text{ }cos\text{ }{{15}^{{}^\circ }}\]
D) \[sin\text{ }{{15}^{{}^\circ }}.cos\text{ }{{75}^{{}^\circ }}\]
Correct Answer: C
Solution :
[c] \[\because sin{{15}^{{}^\circ }}.sin{{15}^{{}^\circ }}\] \[=\frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{{{(\sqrt{3})}^{2}}-{{1}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}}=\frac{2}{4\times 2}=\frac{1}{4}\] which is ratio. Hence, option [c] is correct.
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