Question

Find the value of x if \[{\tan ^{ - 1}}\left( {\dfrac{{x - 3}}{{x - 4}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{x + 3}}{{x + 4}}} \right) = \dfrac{\pi }{4}\].

Answer 1

Hint: Here, we have to solve the given equation and find the value of \[x\]. We will use the formula for the sum of trigonometric inverse of two angles to simplify the equation. Then, you need to solve the equation for \[x\].

Formula Used: We will use the formula \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\] and \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].

Complete step-by-step answer:
We will use the formula for the sum of trigonometric inverse of two angles to simplify the equation.
We know that the sum of trigonometric inverse of two angles \[A\] and \[B\] is given by \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\].
Substituting \[A = \dfrac{{x - 3}}{{x - 4}}\] and \[B = \dfrac{{x + 3}}{{x + 4}}\] in the formula, we get
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{x - 3}}{{x - 4}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{x + 3}}{{x + 4}}} \right) = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 3}}{{x - 4}} + \dfrac{{x + 3}}{{x + 4}}}}{{1 - \left( {\dfrac{{x - 3}}{{x - 4}}} \right)\left( {\dfrac{{x + 3}}{{x + 4}}} \right)}}} \right]\]
Substituting \[{\tan ^{ - 1}}\left( {\dfrac{{x - 3}}{{x - 4}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{x + 3}}{{x + 4}}} \right) = \dfrac{\pi }{4}\] in the equation, we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 3}}{{x - 4}} + \dfrac{{x + 3}}{{x + 4}}}}{{1 - \left( {\dfrac{{x - 3}}{{x - 4}}} \right)\left( {\dfrac{{x + 3}}{{x + 4}}} \right)}}} \right]\]
Now, we will simplify the expression.
Rewriting the expression, we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 3}}{{x - 4}} + \dfrac{{x + 3}}{{x + 4}}}}{{1 - \dfrac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}}}} \right]\]
We know that the product of sum and difference of two numbers is given by the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Using the algebraic identity to simplify the expression, we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 3}}{{x - 4}} + \dfrac{{x + 3}}{{x + 4}}}}{{1 - \dfrac{{{x^2} - 9}}{{{x^2} - 16}}}}} \right]\]
Taking the L.C.M., we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{\left( {x - 3} \right)\left( {x + 4} \right) + \left( {x + 3} \right)\left( {x - 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}}}{{\dfrac{{{x^2} - 16 - {x^2} + 9}}{{{x^2} - 16}}}}} \right]\]
Using the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] again, we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{\left( {x - 3} \right)\left( {x + 4} \right) + \left( {x + 3} \right)\left( {x - 4} \right)}}{{{x^2} - 16}}}}{{\dfrac{{{x^2} - 16 - {x^2} + 9}}{{{x^2} - 16}}}}} \right]\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - 3} \right)\left( {x + 4} \right) + \left( {x + 3} \right)\left( {x - 4} \right)}}{{{x^2} - 16 - {x^2} + 9}}} \right]\\ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {x - 3} \right)\left( {x + 4} \right) + \left( {x + 3} \right)\left( {x - 4} \right)}}{{ - 7}}} \right]\end{array}\]
Multiplying the terms of the expression using the distributive property of multiplication, we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left( {\dfrac{{{x^2} + 4x - 3x - 12 + {x^2} - 4x + 3x - 12}}{{ - 7}}} \right)\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \dfrac{\pi }{4} = {\tan ^{ - 1}}\left( {\dfrac{{2{x^2} - 24}}{{ - 7}}} \right)\]
Now, we know that the equation \[p = {\tan ^{ - 1}}\theta \] can be written as \[\tan p = \theta \].
Therefore, rewriting the equation \[\dfrac{\pi }{4} = {\tan ^{ - 1}}\left( {\dfrac{{2{x^2} - 24}}{{ - 7}}} \right)\], we get
\[ \Rightarrow \tan \dfrac{\pi }{4} = \dfrac{{2{x^2} - 24}}{{ - 7}}\]
Substituting \[\tan \dfrac{\pi }{4} = 1\], we get
\[ \Rightarrow 1 = \dfrac{{2{x^2} - 24}}{{ - 7}}\]
Multiplying both sides by \[ - 7\], we get
\[\begin{array}{l} \Rightarrow 1 \times - 7 = \left( {\dfrac{{2{x^2} - 24}}{{ - 7}}} \right)\left( { - 7} \right)\\ \Rightarrow - 7 = 2{x^2} - 24\end{array}\]
Adding 24 on both sides of the equation, we get
\[\begin{array}{l} \Rightarrow - 7 + 24 = 2{x^2} - 24 + 24\\ \Rightarrow 17 = 2{x^2}\end{array}\]
Dividing both sides by 2, we get
\[ \Rightarrow \dfrac{{17}}{2} = {x^2}\]
Finally, taking the square root of both sides, we get
\[ \Rightarrow x = \pm \sqrt {\dfrac{{17}}{2}} \]
Therefore, the value of \[x\] is \[ \pm \sqrt {\dfrac{{17}}{2}} \].

Note: The equation given to us in the question has inverse trigonometric function. As the name suggests, inverse trigonometric functions are the inverse of basic trigonometric functions. It is also called ‘arc function’ because it gives the length of the arc by performing the opposite operation as a trigonometric function. We have also used the distributive property of multiplication in the solution. The distributive property of multiplication states that \[\left( {a + b} \right)\left( {c + d} \right) = a \cdot c + a \cdot d + b \cdot c + b \cdot d\].