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 :
Putting \[x=\sin \theta \]and \[y=\sin \varphi \] \[\cos \theta +\cos \varphi =a(\sin \theta -\sin \varphi )\] Þ \[2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=a\left\{ 2\cos \frac{\theta +\varphi }{2}\sin \frac{\theta -\varphi }{2} \right\}\] Þ \[\frac{\theta -\varphi }{2}={{\cot }^{-1}}a\Rightarrow \theta -\varphi =2{{\cot }^{-1}}a\] Þ \[{{\sin }^{-1}}x-{{\sin }^{-1}}y=2{{\cot }^{-1}}a\] Þ \[\frac{1}{\sqrt{1-{{x}^{2}}}}-\frac{1}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}\].
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