Question
The value of $$x$$ for which $$\sin \left( {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right) = \cos \left( {{{\tan }^{ - 1}}x} \right)$$ is
A.
$$ \frac{1}{2}$$
B.
1
C.
0
D.
$$ - \frac{1}{2}$$
Answer :
$$ - \frac{1}{2}$$
Solution :
$$\eqalign{
& \sin \left[ {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right) \cr
& \Rightarrow \,\,\sin \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }}} \right)} \right] = \cos \left[ {{{\cos }^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \right] \cr
& \Rightarrow \,\,\frac{1}{{\sqrt {1 + {{\left( {1 + x} \right)}^2}} }} = \frac{1}{{\sqrt {1 + {x^2}} }} \cr
& \Rightarrow \,\,1 + 1 + 2x + {x^2} = 1 + {x^2} \cr
& \Rightarrow \,\,2x + 1 = 0 \cr
& \Rightarrow \,\,x = - \frac{1}{2} \cr} $$