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Question: If \({\tan ^{ - 1}}\dfrac{{X - 3}}{{X - 4}} + {\tan ^{ - 1}}\dfrac{{X + 3}}{{X + 4}} = \dfrac{3}{4},...

If tan⁡−1X−3X−4+tan⁡−1X+3X+4=34,{\tan ^{ - 1}}\dfrac{{X - 3}}{{X - 4}} + {\tan ^{ - 1}}\dfrac{{X + 3}}{{X + 4}} = \dfrac{3}{4}, then find the value of XX .

Explanation

Solution

In this question tan⁡−1X−3X−4+tan⁡−1X+3X+4=34,{\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⁡−1A+tan⁡−1B{\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⁡\tan inverse to the other side in the form of tan⁡\tan .

Step-by-step solution:
Given: tan⁡−1X−3X−4+tan⁡−1X+3X+4=34,{\tan ^{ - 1}}\dfrac{{X - 3}}{{X - 4}} + {\tan ^{ - 1}}\dfrac{{X + 3}}{{X + 4}} = \dfrac{3}{4}, then find the value of XX .
This question is related to inverse trigonometry. So, we know some basic formula of trigonometry like tan⁡−1A+tan⁡−1B{\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⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
Now, ∵tan⁡−1(X−3X−4)+tan⁡−1(X+3X+4)=34\because {\tan ^{ - 1}}\left( {\dfrac{{X - 3}}{{X - 4}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{X + 3}}{{X + 4}}} \right) = \dfrac{3}{4}
⇒tan⁡−1(X−3X−4+X+3X+41−X−3X−4×X+3X+4)=34 ⇒(X−3)(X+4)+(X+3)(X−4)(X−4)(X+4)−(X−3)(X+3)=tan⁡34  \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} \\\
⇒X2+X−12+X2−X−12X2−16−X2+9=tan⁡34\Rightarrow \dfrac{{{X^2} + X - 12 + {X^2} - X - 12}}{{{X^2} - 16 - {X^2} + 9}} = \tan \dfrac{3}{4}
⇒2X2−24−7=tan⁡34⇒2X2=−7tan⁡(34)+24 ∴X2=24−7(tan⁡34)2  \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⁡34\tan \dfrac{3}{4} term without more calculation. So, treat tan⁡(34)\tan \left( {\dfrac{3}{4}} \right) as constant and use a proper inverse trigonometric formula.