Find the value of ${\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + 2{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)$
A. $\dfrac{\pi }{2}$
B. ${\cos ^{ - 1}}\left( {\dfrac{{171}}{{221}}} \right)$
C. $\dfrac{\pi }{4}$
D. None of these

Hint: In the given question, we need to find the value of ${\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + 2{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)$. As we can see we are given an inverse trigonometric function. An inverse trigonometric function is defined as the inverse of the basic trigonometric functions. At first, we will try to convert the inverse function of $\tan $ into the inverse function of $\cos $ by applying $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ formula/identity and simplify the expression. After this we will apply ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right\}$ formula/identity to get our required answer.

 Formula used: $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$

 Complete step by step solution:
We have, ${\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + 2{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)$
As we know that $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$. Therefore, we get
$ = {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\left( {\dfrac{1}{5}} \right)}^2}}}{{1 + {{\left( {\dfrac{1}{5}} \right)}^2}}}} \right)$
$ = {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{1 - \dfrac{1}{{25}}}}{{1 + \dfrac{1}{{25}}}}} \right)$
On taking LCM, we get
$ = {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{\dfrac{{25 - 1}}{{25}}}}{{\dfrac{{25 + 1}}{{25}}}}} \right)$
$ = {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{24}}{{25}} \times \dfrac{{25}}{{26}}} \right)$
On simplification, we get
$ = {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)$
As we know ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right\}$. Therefore, we get
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left\{ {\dfrac{{15}}{{17}} \times \dfrac{{12}}{{13}} - \sqrt {1 - {{\left( {\dfrac{{15}}{{17}}} \right)}^2}} \sqrt {1 - {{\left( {\dfrac{{12}}{{13}}} \right)}^2}} } \right\}$
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left\{ {\dfrac{{180}}{{221}} - \sqrt {1 - \dfrac{{225}}{{289}}} \sqrt {1 - \dfrac{{144}}{{169}}} } \right\}$
On taking LCM, we get
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left\{ {\dfrac{{180}}{{221}} - \sqrt {\dfrac{{289 - 225}}{{289}}} \sqrt {\dfrac{{169 - 144}}{{169}}} } \right\}$
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left\{ {\dfrac{{180}}{{221}} - \sqrt {\dfrac{{64}}{{289}}} \sqrt {\dfrac{{25}}{{169}}} } \right\}$
We can also write the above written expression as
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left\{ {\dfrac{{180}}{{221}} - \sqrt {\dfrac{{{8^2}}}{{{{\left( {17} \right)}^2}}}} \sqrt {\dfrac{{{5^2}}}{{{{\left( {13} \right)}^2}}}} } \right\}$
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{180}}{{221}} - \dfrac{8}{{17}} \times \dfrac{5}{{13}}} \right)$
On multiplication of terms, we get
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{180}}{{221}} - \dfrac{{40}}{{221}}} \right)$
$ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{140}}{{221}}} \right)$
Hence, the value of ${\cos ^{ - 1}}\left( {\dfrac{{15}}{{17}}} \right) + 2{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)$ is ${\cos ^{ - 1}}\left( {\dfrac{{140}}{{221}}} \right)$.

 Therefore, the correct option is D.

 Note: To solve these types of questions, one must remember all the standard formulas of inverse trigonometric functions. Most of the inverse trigonometric functions questions are just based on substitutions, we can only solve the problem if we know the formulas. We should take care of the calculations so as to be sure of our final answer.
Some formulas of inverse trigonometric functions are written below:
1. ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} } \right\}$, if $ - 1 \leqslant x$, $y \leqslant 1$ and ${x^2} + {y^2} \leqslant 1$
2. ${\sin ^{ - 1}}x - {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right\}$, if $ - 1 \leqslant x$, $y \leqslant 1$ and ${x^2} + {y^2} \leqslant 1$
3. ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left\{ {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right\}$, if $ - 1 \leqslant x$, $y \leqslant 1$ and $x + y \geqslant 0$
4. ${\cos ^{ - 1}}x - {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left\{ {xy + \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right\}$, if $ - 1 \leqslant x$, $y \leqslant 1$ and $x \leqslant y$
5. $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$, if $ - 1 < x < 1$
6. $2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$, if $ - 1 \leqslant x \leqslant 1$
7. $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$, if $0 \leqslant x < \infty $
8. $2{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$, if $ - \dfrac{1}{{\sqrt 2 }} \leqslant x \leqslant \dfrac{1}{{\sqrt 2 }}$
9. $2{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {2{x^2} - 1} \right)$, if $0 \leqslant x \leqslant 1$
We use them accordingly to the given problem.