A) 1
B) -1
C) 2
D) None of these
Correct Answer: A
Solution :
| [a] Let \[s={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] and t |
| \[={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] |
| We have to find out \[\frac{ds}{dt};\] putting \[x=\tan \theta ,\] we get |
| \[s={{\sin }^{-1}}\left[ \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right]={{\sin }^{-1}}(\sin 2\theta )=2\theta \] |
| \[=2{{\tan }^{-1}}x\] |
| \[\therefore \frac{ds}{dx}=\frac{2}{1+{{x}^{2}}}\] |
| and \[t={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)={{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\] |
| \[={{\cos }^{-1}}(cos2\theta )=2\theta =2ta{{n}^{-1}}x\] |
| \[\therefore \frac{dt}{dx}=\frac{2}{1+{{x}^{2}}}\] |
| \[\therefore \frac{ds}{dt}=\frac{ds/dx}{dt/dx}=\frac{2}{1+{{x}^{2}}}\times \frac{1+{{x}^{2}}}{2}=1\] |
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