Find the range of $f (x) = {sin}^{- 1} x + {cos}^{- 1} x + {tan}^{- 1} x$ .

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Solution

${tan}^{- 1} x$ is defined for $x \in (- \infty, \infty)$

${sin}^{- 1} x + {cos}^{- 1} x$ are defined for $x \in [- 1, 1]$

so, $f (x)$ ${sin}^{- 1} x + {cos}^{- 1} x + {tan}^{- 1} x$ is defined for $x \in [- 1, 1]$

for $x \in [- 1, 1]$ , ${sin}^{- 1} x + {cos}^{- 1} x = π / 2$

and $- \frac{π}{4} \leq {tan}^{- 1} x \leq \frac{π}{4}$

$\Rightarrow$ $\frac{π}{2} - \frac{π}{4} \leq {sin}^{- 1} x + {cos}^{- 1} x + {tan}^{- 1} x \leq \frac{π}{2} + \frac{π}{4}$

$\Rightarrow$ Range = $[\frac{π}{4}, \frac{3 π}{4}]$

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Similar questions

f (x) = {tan}^{- 1} [\frac{2}{π} (2 {tan}^{- 1} x - {sin}^{- 1} x + {cot}^{- 1} x - {cos}^{- 1} x)]

f (x) = {sin}^{- 1} x + {tan}^{- 1} x + {sec}^{- 1} x

f (x) = {tan}^{- 1} x + {sin}^{- 1} x + {cos}^{- 1} x

(0, π)

f (x) = {tan}^{- 1} x + {sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2} + {tan}^{- 1} x

x ϵ (- 1, 1]

f (x) = {sin}^{- 1} x - {cos}^{- 1} x

	List I		List II
A.	Range of $f (x) = {sin}^{- 1} x + {cos}^{- 1} x + {cot}^{- 1} x$ is	1.	$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
B.	Range of $f (x) = cot - 1 x + {tan}^{- 1} x + c o s e c^{- 1} x$ is	2.	$[\frac{π}{2}, \frac{3 π}{2}]$
C.	Range of $f (x) = {cot}^{- 1} x + {tan}^{- 1} x + {cos}^{- 1} x$ is	3.	${0, π}$
D.	Range of $f (x) = {sec}^{- 1} x + c o s e c^{- 1} x + {sin}^{- 1} x$ is	4.	$(\frac{π}{2}, \frac{3 π}{2})$

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