A) \[x=1,y=2;x=2,y=7\]
B) \[x=1,y=3;x=2,y=4\]
C) \[x=0,y=0;x=3,y=4\]
D) None of these
Correct Answer: A
Solution :
[a] Converting cos and sin into tan, we get, \[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \frac{1}{y} \right)={{\tan }^{-1}}\left( \frac{3}{1} \right)\] \[\Rightarrow {{\tan }^{-1}}\left[ \frac{x+(1/y)}{1-x(1/y)} \right]={{\tan }^{-1}}3\] \[\Rightarrow {{\tan }^{-1}}\left( \frac{xy+1}{y-x} \right)={{\tan }^{-1}}3\Rightarrow \frac{xy+1}{y-x}=3\] \[\Rightarrow x=\frac{3y-1}{y+3}=3-\frac{10}{y+3}\therefore <3\] So, for \[x=1,\,\,y=2\] and for \[z=2,\,\,y=7\]
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