$\sin^{-1}(\frac{1+x^2}{2x})$ is
Answer & explanation
Correct answer: option 3
$\sin^{-1}(\frac{1+x^2}{2x})$ is defined only for x = – 1 and x = 1.
Hence (C) is the correct answer.
$\sin^{-1}(\frac{1+x^2}{2x})$ is
Correct answer: option 3
$\sin^{-1}(\frac{1+x^2}{2x})$ is defined only for x = – 1 and x = 1.
Hence (C) is the correct answer.