If \[y={{\sec }^{-1}}\left( \frac{x+1}{x-1} \right)+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\], then \[\frac{dy}{dx}=\] [MNR 1984]

If \[y={{\sec }^{-1}}\left( \frac{x+1}{x-1} \right)+{{\sin }^{-1}}\lef

1.	If \[y={{\sec }^{-1}}\left( \frac{x+1}{x-1} \right)+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\], then \[\frac{dy}{dx}=\] [MNR 1984]
A.	0
B.	1
C.	2
D.	3
Answer» B. 1

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If \[y={{\sec }^{-1}}\left( \frac{x+1}{x-1} \right)+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\], then \[\frac{dy}{dx}=\] [MNR 1984]

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