A) \[\sqrt{\frac{1-{{x}^{2}}}{1-{{y}^{2}}}}\]
B) \[\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}\]
C) \[\sqrt{\frac{{{x}^{2}}-1}{1-{{y}^{2}}}}\]
D) \[\sqrt{\frac{{{y}^{2}}-1}{1-{{x}^{2}}}}\]
Correct Answer: B
Solution :
[b] Putting \[x=\sin \theta \]and \[y=\sin \phi \] \[\cos \theta +\cos \phi =a(sin\theta -sin\phi )\] \[\Rightarrow 2\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}=a\left\{ 2\cos \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2} \right\}\] \[\Rightarrow \frac{\theta -\phi }{2}={{\cos }^{-1}}a\] \[\Rightarrow \theta -\phi =2{{\cot }^{-1}}a\] \[\Rightarrow {{\sin }^{-1}}x-{{\sin }^{-1}}y=2{{\cot }^{-1}}a\] \[\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}-\frac{1}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}\]
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