QJEEAdv08Ma7b8f: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma7b8f Solution: If , then

Given that,

0 < x < 1 ,

$\sqrt {1 + {x^2}} {[{\{ x\cos ({\cot ^{ - 1}}x) + \sin ({\cot ^{ - 1}}x)\} ^2} - 1]^{{1 \over 2}}}$

Find : value of this expression

Here, 0 < x < 1 ,

${\cot ^{ - 1}}x = {\sin ^{ - 1}}\left( {{1 \over {\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {{x \over {\sqrt {1 + {x^2}} }}} \right)$

Now, $\sqrt {1 + {x^2}} {[{\{ x\cos ({\cot ^{ - 1}}x) + \sin ({\cot ^{ - 1}}x)\} ^2} - 1]^{{1 \over 2}}}$

$\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cos }^{ - 1}}\left( {{x \over {1 + {x^2}}}} \right)} \right) + \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt {1 + {x^2}} }}} \right)} \right\}}^2}} \right]^{{1 \over 2}}}$

$= \sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\,.\,{x \over {\sqrt {1 + {x^2}} }} + {1 \over {\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right]^{{1 \over 2}}}$

$= \sqrt {1 + {x^2}} {\left[ {{{\left\{ {{{{x^2} + 1} \over {\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right]^{{1 \over 2}}}$

$= \sqrt {1 + {x^2}} {[{x^2} + 1 - 1]^{{1 \over 2}}}$

$= x\sqrt {1 + {x^2}}$

which is the value of the expression.