Question

If ${\tan ^{ - 1}}\dfrac{{X - 3}}{{X - 4}} + {\tan ^{ - 1}}\dfrac{{X + 3}}{{X + 4}} = \dfrac{3}{4},$ then find the value of $X$ .

Answer 1

Hint: In this question ${\tan ^{ - 1}}\dfrac{{X - 3}}{{X - 4}} + {\tan ^{ - 1}}\dfrac{{X + 3}}{{X + 4}} = \dfrac{3}{4},$ to solve this question we have to apply the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B$ . Using this formula, the given question can be solved in the proper way. So, proceed to the solution with this formula. When we perform the operations, we have to carry $\tan $ inverse to the other side in the form of $\tan $ .

Step-by-step solution:
Given: ${\tan ^{ - 1}}\dfrac{{X - 3}}{{X - 4}} + {\tan ^{ - 1}}\dfrac{{X + 3}}{{X + 4}} = \dfrac{3}{4},$ then find the value of $X$ .
This question is related to inverse trigonometry. So, we know some basic formula of trigonometry like ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B$ . So, using the basic formula we will get some relevant terms that can be solved easily.
Formula used: ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)$
Now, $\because {\tan ^{ - 1}}\left( {\dfrac{{X - 3}}{{X - 4}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{X + 3}}{{X + 4}}} \right) = \dfrac{3}{4}$
$
   \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{X - 3}}{{X - 4}} + \dfrac{{X + 3}}{{X + 4}}}}{{1 - \dfrac{{X - 3}}{{X - 4}} \times \dfrac{{X + 3}}{{X + 4}}}}} \right) = \dfrac{3}{4} \\
   \Rightarrow \dfrac{{(X - 3)(X + 4) + (X + 3)(X - 4)}}{{(X - 4)(X + 4) - (X - 3)(X + 3)}} = \tan \dfrac{3}{4} \\
 $
$ \Rightarrow \dfrac{{{X^2} + X - 12 + {X^2} - X - 12}}{{{X^2} - 16 - {X^2} + 9}} = \tan \dfrac{3}{4}$
$
   \Rightarrow \dfrac{{2{X^2} - 24}}{{ - 7}} = \tan \dfrac{3}{4} \Rightarrow 2{X^2} = - 7\tan \left( {\dfrac{3}{4}} \right) + 24 \\
  \therefore {X^2} = \dfrac{{24 - 7\left( {\tan \dfrac{3}{4}} \right)}}{2} \\
 $
Note: In this question there are some irrelevant values given but as a student we can proceed the $\tan \dfrac{3}{4}$ term without more calculation. So, treat $\tan \left( {\dfrac{3}{4}} \right)$ as constant and use a proper inverse trigonometric formula.