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 \[sin\alpha =xsin\beta \Rightarrow sin\beta =\frac{1}{x}sin\alpha \] \[\Rightarrow cosec\beta =\frac{x}{{{\tan }^{2}}\alpha }\] \[[\therefore cose{{c}^{2}}\beta -co{{t}^{2}}\beta =1]\] \[\Rightarrow \frac{{{x}^{2}}}{{{\sin }^{2}}\alpha }-\frac{{{y}^{2}}}{{{\tan }^{2}}\alpha }=1\] \[\Rightarrow \frac{{{x}^{2}}}{{{\sin }^{2}}\alpha }-\frac{{{y}^{2}}{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }=1\] \[{{x}^{2}}-\text{ }{{y}^{2}}co{{s}^{2}}\alpha =si{{n}^{2}}\alpha \] \[=1-co{{s}^{2}}\alpha \] \[\Rightarrow {{x}^{2}}-1={{y}^{2}}co{{s}^{2}}\alpha -co{{s}^{2}}a\] \[=\left( {{y}^{2}}-1 \right){{\cos }^{2}}\alpha \] \[\Rightarrow co{{s}^{2}}\alpha =\frac{{{x}^{2}}-1}{{{y}^{2}}-1}\]
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