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# Find the value of $\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right) + \tan \left( {\dfrac{\pi }{4} - \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)$$\left( A \right).{\text{ }}\dfrac{{2a}}{b} \\ \left( B \right).{\text{ }}\dfrac{a}{b} \\ \left( C \right).{\text{ }}\dfrac{b}{a} \\ \left( D \right).{\text{ }}\dfrac{{2b}}{a} \\$ Verified
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Hint: Solve by using simple trigonometric identities of $\tan \theta$ and $\cos \theta$.

Given $\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right) + \tan \left( {\dfrac{\pi }{4} - \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right){\text{ }} \ldots \left( 1 \right)$
Let $\dfrac{1}{2}{\cos ^{ - 1}}\dfrac{a}{b} = \theta$
$\therefore {\cos ^{ - 1}}\dfrac{a}{b} = 2\theta \\ \cos 2\theta = \dfrac{a}{b}{\text{ }} \ldots \left( 2 \right) \\$
Put the value of $\dfrac{1}{2}{\cos ^{ - 1}}\dfrac{a}{b} = \theta$ in equation $\left( 1 \right)$, we get
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) + \tan \left( {\dfrac{\pi }{4} - \theta } \right){\text{ }} \ldots \left( 3 \right)$
We know that, $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$and $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
Using above identities in$\left( 3 \right)$, we get
$\Rightarrow \dfrac{{\tan \dfrac{\pi }{4} + \tan \theta }}{{1 - \tan \dfrac{\pi }{4}\tan \theta }} + \dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4}\tan \theta }}$
Putting the value of $\tan \dfrac{\pi }{4} = 1$ in above equation, we get
$\Rightarrow \dfrac{{1 + \tan \theta }}{{1 - 1\tan \theta }} + \dfrac{{1 - \tan \theta }}{{1 + 1\tan \theta }} \\ \Rightarrow \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} + \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }} \\$
Using cross multiplication, we get
$\Rightarrow \dfrac{{{{\left( {1 + \tan \theta } \right)}^2} + {{\left( {1 - \tan \theta } \right)}^2}}}{{\left( {1 - \tan \theta } \right)\left( {1 + \tan \theta } \right)}} \\ \Rightarrow \dfrac{{1 + 2\tan \theta + {{\tan }^2}\theta + 1 - 2\tan \theta + {{\tan }^2}\theta }}{{{{\left( 1 \right)}^2} - {{\left( {\tan \theta } \right)}^2}}}{\text{ }}\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}{\text{ and }}{{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right] \\ \Rightarrow \dfrac{{2 + 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }} \\$
Taking $2$ common from numerator, we get
$\Rightarrow 2\left( {\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}} \right)$
We know that $\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$. Hence, we can write above equation as:
$\Rightarrow \dfrac{2}{{\left( {\cos 2\theta } \right)}}$
Now, putting the value of $\cos 2\theta$ from equation$\left( 2 \right)$in above equation, we get
$\Rightarrow \dfrac{2}{{\left( {\dfrac{a}{b}} \right)}} \\ \Rightarrow \dfrac{{2b}}{a} \\$
$\therefore$Correct option is $\left( D \right)$.

Note: In these types of problems, one should always try to convert the equation to some trigonometric identity by either taking out the common terms from the equations or try to minimize it by using the identities.