A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
| We have, \[1+\cos 2X-2\cos 2Y=2\sin X\sin Y\] |
| \[\Rightarrow 1+1-2{{\sin }^{2}}X-2\left( 1-2{{\sin }^{2}}y \right)\]\[=2\sin X\sin Y\]\[\Rightarrow 4{{\sin }^{2}}Y-2{{\sin }^{2}}X-2\sin X\sin Y=0\]\[\Rightarrow \,\,4{{b}^{2}}-2{{a}^{2}}-2ab=0\] |
| \[\left[ \therefore \frac{a}{\sin X}=\frac{a}{\sin Y}=\frac{a}{\sin Z}=k \right]\]\[\Rightarrow \frac{{{a}^{2}}}{{{b}^{2}}}+\frac{a}{b}-2=0\]\[\Rightarrow {{\left( \frac{a}{b} \right)}^{2}}+\left( \frac{a}{b} \right)-2=0\] |
| \[\therefore \,\,\frac{a}{b}=1\] |
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