Find the value of \[\sin \left( {4{{\tan }^{ - 1}}\dfrac{1}{3}} \right)\] .
A.\[\dfrac{{12}}{{25}}\]
B. \[\dfrac{{24}}{{25}}\]
C. \[\dfrac{1}{5}\]
D. None of these

Hint:Here the given function is of sine and tangent. So first use the formula of \[2{\tan ^{ - 1}}x\] in term of \[{\tan ^{ - 1}}x\] then again use the formula of \[2{\tan ^{ - 1}}x\] in term of \[{\sin ^{ - 1}}x\] then calculate to obtain the required result.

Formula used
1) \[2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2\cdot x}}{{1 - {x^2}}}} \right), - 1 < x < 1\]
2) \[2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{2\cdot x}}{{1 + {x^2}}}} \right),\left| x \right| \le 1\]
3) \[\sin ({\sin ^{ - 1}}x) = x\]

Complete step by step solution
The given expression is,
\[\sin \left( {4{{\tan }^{ - 1}}\dfrac{1}{3}} \right)\]
\[ = \sin \left( {2\cdot2{{\tan }^{ - 1}}\dfrac{1}{3}} \right)\]
Apply the formula \[2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2\cdot x}}{{1 - {x^2}}}} \right)\]in the expression \[2{\tan ^{ - 1}}\dfrac{1}{3}\] for further calculation.
\[ = \sin \left( {2\cdot{{\tan }^{ - 1}}\dfrac{{2\cdot\dfrac{1}{3}}}{{1 - {{\left( {\dfrac{1}{3}} \right)}^2}}}} \right)\]
\[ = \sin \left( {2\cdot{{\tan }^{ - 1}}\dfrac{{\dfrac{2}{3}}}{{\dfrac{8}{9}}}} \right)\]
\[ = \sin \left( {2\cdot{{\tan }^{ - 1}}\dfrac{3}{4}} \right)\]
Apply the formula \[2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{2\cdot x}}{{1 + {x^2}}}} \right)\]in the expression \[2{\tan ^{ - 1}}\dfrac{3}{4}\] for further calculation.
\[ = \sin \left( {{{\sin }^{ - 1}}\dfrac{{2\cdot\dfrac{3}{4}}}{{1 + {{\left( {\dfrac{3}{4}} \right)}^2}}}} \right)\]
\[ = \sin \left( {{{\sin }^{ - 1}}\dfrac{6}{4}\cdot \dfrac{{16}}{{25}}} \right)\]
\[ = \sin \left( {{{\sin }^{ - 1}}\dfrac{{24}}{{25}}} \right)\]
\[ = \dfrac{{24}}{{25}}\]
So, the required value of the given expression is \[\dfrac{{24}}{{25}}\].
The correct option is B.

Additional information
We are converting inverse tangent to inverse tangent first then according to the demand of the question we then convert inverse tangent to the inverse sine function, if the question contains the cosine function then we can also convert it to inverse cosine function as there is a formula by which we can convert inverse tangent function to inverse cosine function also.

Note Sometimes students use calculators and just place the values to obtain the required result but in exams, we cannot use calculators so remember all the formulas used in this problem. In this type of problem, we use two types of formulas for one expression as per requirement.