Question
What is $$\sin \left[ {{{\cot }^{ - 1}}\left\{ {\cos \left( {{{\tan }^{ - 1}}x} \right)} \right\}} \right]$$ where $$x > 0,$$ equal to ?
A.
$$\sqrt {\frac{{\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 2} \right)}}} $$
B.
$$\sqrt {\frac{{\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 1} \right)}}} $$
C.
$${\frac{{\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 2} \right)}}}$$
D.
$${\frac{{\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 1} \right)}}}$$
Answer :
$$\sqrt {\frac{{\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 2} \right)}}} $$
Solution :
$$\eqalign{
& {\text{Let, }}\alpha = {\tan ^{ - 1}}x \cr
& \Rightarrow \tan \alpha = x \cr
& {\text{then }}\,\cos \alpha = \frac{1}{{\sqrt {1 + {{\tan }^2}\alpha } }} = \frac{1}{{\sqrt {1 + {x^2}} }} \cr
& \Rightarrow \cos \left. {\left( {{{\tan }^{ - 1}}x} \right)} \right\} = \left\{ {\frac{1}{{\sqrt {1 + {x^2}} }}} \right\} \cr
& {\text{So}},\,\,{\cot ^{ - 1}}\cos \left( {{{\tan }^{ - 1}}x} \right) = {\cot ^{ - 1}}\left\{ {\frac{1}{{\sqrt {1 + {x^2}} }}} \right\} \cr
& {\text{Let, }}{\cot ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right) = \beta \cr
& \Rightarrow \cot \beta = \frac{1}{{\sqrt {1 + {x^2}} }} \cr
& {\text{and }}\,\sin \beta = \frac{1}{{\sqrt {1 + {{\cot }^2}\beta } }} \cr
& = \frac{{\sqrt {1 + {x^2}} }}{{\sqrt {{x^2} + 1 + 1} }} = \sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}} \cr
& \Rightarrow \sin \left[ {{{\cot }^{ - 1}}\left\{ {\cos \left( {{{\tan }^{ - 1}}} \right)} \right\}} \right] = \sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}} \cr} $$