A) 0
B) \[\text{tan}\,t\]
C) 1
D) \[\text{sin}\,t\text{ cos}\,t\]
Correct Answer: C
Solution :
Given, \[x={{\cos }^{-1}}\left( \frac{1}{\sqrt{1+{{t}^{2}}}} \right)\] and \[y={{\sin }^{-1}}\left( \frac{t}{\sqrt{1+{{t}^{2}}}} \right)\] \[\Rightarrow \] \[x={{\tan }^{-1}}t\] and \[y={{\tan }^{-1}}t\] \[\therefore \] \[y=x\] \[\Rightarrow \] \[\frac{dy}{dx}=1\]You need to login to perform this action.
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