A) \[\cos \beta \]
B) \[\cos 2\beta \]
C) \[\sin \alpha \]
D) \[\sin 2\alpha \]
Correct Answer: B
Solution :
\[\text{1}00\text{1}=\text{91}\times \text{1}0+\text{91}\] \[\text{91}0=\text{91}\times \text{1}0+0\]\[\therefore \] \[=\left( \frac{144}{48}+\frac{384}{48}+\frac{240}{48} \right)=3+8+5=16\]\[\frac{a}{b}\] \[\frac{c}{d}=\frac{L.C.M.(a,c)}{H.C.F.(b,d)}\] \[\Rightarrow \]
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