A) \[\frac{{{x}^{2}}}{{{y}^{2}}+1}\]
B) \[\frac{{{x}^{2}}}{{{y}^{2}}}\]
C) \[\frac{{{x}^{2}}-1}{{{y}^{2}}-1}\]
D) \[\frac{{{x}^{2}}+1}{{{y}^{2}}+1}\]
Correct Answer: C
Solution :
(c): \[tan\alpha =ytan\beta \] \[\Rightarrow tan\beta =\frac{1}{y}tan\alpha \] \[\Rightarrow cot\beta =\frac{y}{\tan \alpha }and\] \[sina=x\,sin\beta \Rightarrow sin\beta =\frac{1}{x}sin\alpha \] \[\Rightarrow \text{cosec}\beta =\frac{x}{\sin \alpha }\] \[\left[ \therefore \cos e{{c}^{2}}\beta -{{\cot }^{2}}\beta =1 \right]\] \[\Rightarrow \frac{{{x}^{2}}}{{{\sin }^{2}}\alpha }-\frac{{{y}^{2}}}{{{\tan }^{2}}\alpha }=1\] \[\Rightarrow \frac{{{x}^{2}}}{{{\sin }^{2}}\alpha }-\frac{{{y}^{2}}{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }=1\] \[\Rightarrow {{x}^{2}}-{{y}^{2}}{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha \] \[=1-{{\cos }^{2}}\alpha \] \[\Rightarrow {{x}^{2}}-1={{y}^{2}}{{\cos }^{2}}\alpha -{{\cos }^{2}}\alpha \] \[=\left( {{y}^{2}}-1 \right){{\cos }^{2}}\alpha \] \[\Rightarrow {{\cos }^{2}}\alpha \frac{{{x}^{2}}-1}{{{y}^{2}}-1}\]
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