Question

Find the domain of the following:$cosec^-1x+cot^-1x+tan^-1x$

Answer 1

$\large\underline{\sf{Solution-}}$

Given

$\rm :\longmapsto\: {cosec}^{ - 1}x + {cot}^{ - 1}x + {tan}^{ - 1}x$

We know,

$\rm :\longmapsto\: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2}$

Therefore,

$\rm :\longmapsto\: {cosec}^{ - 1}x + {cot}^{ - 1}x + {tan}^{ - 1}x$

$\rm \: = \: \: \: {cosec}^{ - 1}x + \dfrac{\pi}{2}$

Now,

We know,

$\rm :\longmapsto\: {cosec}^{ - 1}x \: \in \: \bigg[ - \dfrac{\pi}{2} ,0\bigg) \cup\bigg(0, \dfrac{\pi}{2} \bigg]$

So,

$\rm :\longmapsto\: \dfrac{\pi}{2} + {cosec}^{ - 1}x \: \in \: \bigg[ - \dfrac{\pi}{2} +\dfrac{\pi}{2},0 + \dfrac{\pi}{2}\bigg) \cup\bigg(0 +\dfrac{\pi}{2} , \dfrac{\pi}{2} + \dfrac{\pi}{2} \bigg]$

$\rm :\longmapsto\: \dfrac{\pi}{2} + {cosec}^{ - 1}x \: \in \: \bigg[ 0, \dfrac{\pi}{2}\bigg) \cup\bigg(\dfrac{\pi}{2} , \pi \bigg]$

Hence,

★ Domain of

$\rm :\longmapsto\:{cosec}^{- 1}x + {tan}^{ - 1}x+{cot}^{-1}x\: \in \: \bigg[ 0, \dfrac{\pi}{2}\bigg) \cup\bigg(\dfrac{\pi}{2} , \pi \bigg]$

Additional information :-

$\rm :\longmapsto\: {sin}^{ - 1}x \: \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg]$

$\rm :\longmapsto\: {tan}^{ - 1}x \: \in \: \bigg( - \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg)$

$\rm :\longmapsto\: {cos}^{ - 1}x = \bigg[0, \: \pi\bigg]$

$\rm :\longmapsto\: {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2}$

$\rm :\longmapsto\: {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2}$