A) \[\frac{2\,t}{{{t}^{2}}+1}\]
B) \[\frac{2\,t}{{{t}^{2}}-1}\]
C) \[\frac{2\,t}{1-{{t}^{2}}}\]
D) None of these
Correct Answer: B
Solution :
\[x=\frac{2t}{1+{{t}^{2}}},\,\,\,y=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\] Put \[t=\tan \theta \] \[x=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta \], \[y=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta \] \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-2\sin 2\theta }{2\cos 2\theta }\] = \[-\tan 2\theta \]= \[\frac{-2\tan \theta }{1-{{\tan }^{2}}\theta }\] = \[\frac{-2t}{1-{{t}^{2}}}\]= \[\frac{2t}{{{t}^{2}}-1}\].
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