Range of f(x) =
${sin}^{- 1} x + {tan}^{- 1} x + {sec}^{- 1} x is$

(\frac{π}{4}, \frac{3 π}{4})

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(\frac{π}{4}, \frac{3 π}{4}]

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{\frac{π}{4}, \frac{3 π}{4}}

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none of these

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Solution

The correct option is C ${\frac{π}{4}, \frac{3 π}{4}}$
$Domain of {sin}^{- 1} x i s (- 1, 1] .$
$Domain of {tan}^{- 1} x i s R .$
$Domain of {sec}^{- 1} x i s$
$(- \infty, - 1] \cup [1, \infty) .$
So, domain of
${sin}^{- 1} x + {tan}^{- 1} x + {sec}^{- 1} x is the intersection of all these three domains i . e ., (- 1, 1} .$
$For x = - 1, {sin}^{- 1} x + {tan}^{- 1} x + {sec}^{- 1} x = {sin}^{- 1} (- 1) + {tan}^{- 1} (- 1) + {sec}^{- 1} (- 1) = - \frac{π}{2} - \frac{π}{4} + π = \frac{π}{4} .$
$Similarly, For x = 1, {sin}^{- 1} x + {tan}^{- 1} x + {sec}^{- 1} x = \frac{π}{2} + \frac{π}{4} + 0 = \frac{3 π}{4} .$
Therefore, the range is
$(\frac{π}{4}, \frac{3 π}{4}} .$

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