Question

The value of $\tan \left[ {\dfrac{\pi }{4} + \left( {\dfrac{1}{2}} \right){{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right] + \tan \left[ {\dfrac{\pi }{4} - \left( {\dfrac{1}{2}} \right){{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right]$ is equal to
A. $\dfrac{{2a}}{b}$
B. $\dfrac{{2b}}{a}$
C. $\dfrac{a}{b}$
D. $\dfrac{b}{a}$

Answer 1

Hint: In the given question, we need to find the value of the given function. For that, we apply the trigonometric formulas in the given question and then simplify it to get the desired result.

Formula Used:
We have been using the following formulas:
1. $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\,\tan B}}$
2. $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
3. $\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$
4. ${\sec ^2}A = 1 + {\tan ^2}A$
5. The value of $\tan \left( {\dfrac{\pi }{4}} \right) = 1$
6. ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
7. ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

Complete step by step solution:
 Given function is $\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right) + \tan \left( {\dfrac{\pi }{4} - \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)$
Now we know that $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\,\tan B}}$ and $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$.So, by applying the formula in given function, we get
$\dfrac{{\tan \dfrac{\pi }{4} + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}}{{1 - \tan \dfrac{\pi }{4}\tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}} + \dfrac{{\tan \dfrac{\pi }{4} - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}}{{1 + \tan \dfrac{\pi }{4}\tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}}$
Now we know that $\tan \left( {\dfrac{\pi }{4}} \right) = 1$
So, our function becomes:
$\dfrac{{1 + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}}{{1 - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}} + \dfrac{{1 - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}}{{1 + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)}}$
Now we take L.C.M, we get
$\dfrac{{\left[ {\left( {1 + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)} \right)\,\,\,\left( {1 + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)} \right)} \right] + \left[ {\left( {1 - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)} \right)\left( {1 - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)} \right)} \right]}}{{\left( {1 - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)} \right)\left( {1 + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)} \right)}}$
By simplifying, we get
$\dfrac{{{{\left[ {1 + \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)} \right]}^2}\, + \,{{\left[ {1 - \tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)} \right]}^2}}}{{1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}\left( {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right)$
Now we know that ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
Therefore, our function becomes
$\dfrac{{1 + {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) + 2\tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) + 1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) - 2\tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}$
By simplifying, we get
$
  \dfrac{{1 + {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) + 2\tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) + 1 + {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) - 2\tan \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}} \\
  \dfrac{{1 + {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right) + 1 + {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}} \\
  \dfrac{{2 + 2{{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}} \\
  \dfrac{{2\left( {1 + {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}} \\
 $
Now we know that $\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$
Now by applying this formula in above function, we get
$\dfrac{2}{{\cos 2\,\left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right)}}$
Now we know that $\cos \left( {{{\cos }^{ - 1}}x} \right) = x$
Therefore, our function becomes:
$\dfrac{2}{{\dfrac{a}{b}}}$
Further simplification, we get
$\dfrac{{2b}}{a}$
Therefore, the value of $\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right) + \tan \left( {\dfrac{\pi }{4} - \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)$ is equal to $\dfrac{{2b}}{a}$

Option ‘B’ is correct

Note: To solve this type of question, we need to remember all the trigonometric formulas like ${\sec ^2}A = 1 + {\tan ^2}A$ and the algebraic identity like $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ and students should be very careful while applying the formula and simplification.