A) \[0\]
B) \[1\]
C) \[-1\]
D) \[\infty \]
Correct Answer: B
Solution :
Given,\[y={{\sin }^{-1}}\left( \frac{2t}{1+{{t}^{2}}} \right),\,\,x={{\tan }^{-1}}\left( \frac{2t}{1-{{t}^{2}}} \right)\] Let \[t=\tan (\phi /2)\] \[\therefore \] \[\frac{2t}{1+{{t}^{2}}}=\frac{2\tan (\phi /2)}{1+{{\tan }^{2}}(\phi /2)}\] \[=2(\phi /2)=\sin \phi \] \[\Rightarrow \] \[\frac{2t}{1+{{t}^{2}}}=\sin \phi \] ? (i) Also, \[\frac{2t}{1-{{t}^{2}}}=\frac{2\tan (\phi /2)}{1-{{\tan }^{2}}(\phi /2)}\] \[=\tan 2(\phi /2)=\tan \phi \] ? (ii) From (i) and (ii), \[y={{\sin }^{-1}}\sin \phi \]and\[x={{\tan }^{-1}}\tan (\phi )\] \[\Rightarrow \] \[y=\phi ,\,\,x=\phi \Rightarrow \frac{dy}{d\phi }=1,\,\,\frac{dx}{d\phi }=1\] \[\therefore \] \[\frac{dy}{dx}=1\] Alternate Method: Given, \[y={{\sin }^{-1}}\left( \frac{2t}{1+{{t}^{2}}} \right),\,\,x={{\tan }^{-1}}\left( \frac{2t}{1-{{t}^{2}}} \right)\] \[\Rightarrow \] \[\frac{dy}{dt}=\frac{1}{\sqrt{1-{{\left( \frac{2t}{1+{{t}^{2}}} \right)}^{2}}}}\times \frac{d}{dt}\left( \frac{2t}{1+{{t}^{2}}} \right)\] \[=\frac{1+{{t}^{2}}}{1-{{t}^{2}}}\left[ \frac{{{(1+t)}^{2}}\times 2-2t(2t)}{{{(1+{{t}^{2}})}^{2}}} \right]=\frac{2}{1+{{t}^{2}}}\] ? (i) and \[\frac{dx}{dt}=\frac{1}{1+{{\left( \frac{2t}{1-{{t}^{2}}} \right)}^{2}}}\frac{d}{dt}\left( \frac{2t}{1-{{t}^{2}}} \right)\] \[=\frac{1}{1+\left( \frac{2t}{1-{{t}^{2}}} \right)}\frac{(1-{{t}^{2}})\times 2+2t(2t)}{{{(1-{{t}^{2}})}^{2}}}\] \[=\frac{{{(1-{{t}^{2}})}^{2}}}{{{(1+{{t}^{2}})}^{2}}}\left[ \frac{(1+{{t}^{2}})}{{{(1-{{t}^{2}})}^{2}}} \right]=\frac{2}{1+{{t}^{2}}}\] ? (ii) \[\therefore \]\[\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{2}{1+{{t}^{2}}}\times \frac{1+{{t}^{2}}}{2}=1\] [from (i) and (ii)]
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