A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
| [b] \[{{x}^{2}}+{{y}^{2}}+3={{(cosec\theta -sin\theta )}^{2}}+{{(sec\theta -cos\theta )}^{2}}+3\] \[=\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta -3+3\] \[=\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\] \[{{x}^{2}}{{y}^{2}}={{(cosec\theta -sin\theta )}^{2}}{{(sec\theta -cos\theta )}^{2}}\] \[={{\left( \frac{1}{\sin \theta }-\sin \theta \right)}^{2}}{{\left( \frac{1}{\cos \theta }-\cos \theta \right)}^{2}}\] \[={{\left( \frac{1-{{\sin }^{2}}\theta }{\sin \theta }.\frac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)}^{2}}\] \[={{\left( \frac{{{\cos }^{2}}\theta .{{\sin }^{2}}\theta }{\sin \theta \cos \theta } \right)}^{2}}\] \[={{\cos }^{2}}\theta {{\sin }^{2}}\theta \] \[\therefore \] \[{{x}^{2}}{{y}^{2}}({{x}^{2}}+{{y}^{2}}+3)\] \[=\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\cdot {{\cos }^{2}}\theta {{\sin }^{2}}\theta =1\] |
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