Question

Solve the given inverse trigonometric equation and find the value of x.
${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4}$

Answer 1

Hint: We will apply a trigonometric formula which is given by ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$. We will use this formula in order to solve the problem.

Complete step by step solution:
We will consider the expression ${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4}...(i)$. Now we will take the left hand side of equation (i). After this we will apply the formula which is given by ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$. Therefore, we have
$\begin{align}
  & {{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)={{\tan }^{-1}}\left( \dfrac{2x+3x}{1-\left( 2x \right)\left( 3x \right)} \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)={{\tan }^{-1}}\left( \dfrac{5x}{1-\left( 6{{x}^{2}} \right)} \right) \\
\end{align}$
Now as we have ${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4}$ in equation (i). Therefore, we can have ${{\tan }^{-1}}\left( \dfrac{5x}{1-\left( 6{{x}^{2}} \right)} \right)=\dfrac{\pi }{4}$. By taking the tan operation to the right side of the equation we will have $\left( \dfrac{5x}{1-\left( 6{{x}^{2}} \right)} \right)=\tan \left( \dfrac{\pi }{4} \right)$. As we know that the value of $\tan \left( \dfrac{\pi }{4} \right)=1$. Therefore, we will have
$\begin{align}
  & \left( \dfrac{5x}{1-\left( 6{{x}^{2}} \right)} \right)=1 \\
 & \Rightarrow \dfrac{5x}{1-\left( 6{{x}^{2}} \right)}=1 \\
 & \Rightarrow 5x=1-6{{x}^{2}} \\
 & \Rightarrow 6{{x}^{2}}+5x-1=0 \\
\end{align}$
Now we will apply the formula of square root to find the roots of the quadratic equation. The formula is given by
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 6 \right)\left( -1 \right)}}{2\left( 6 \right)} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{25+24}}{12} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{49}}{12} \\
 & \Rightarrow x=\dfrac{-5\pm 7}{12} \\
\end{align}$
Therefore we got two roots here which are given by $x=\dfrac{-5+7}{12}$ and $x=\dfrac{-5-7}{12}$. After simplifying we have the roots as $x=\dfrac{2}{12}$ and $x=\dfrac{-12}{12}$. Thus we have x = $\dfrac{1}{6}$ and x = - 1.
Since, we have got two roots so, we will substitute each values of x in the equation (i). So, we will substitute x = -1 in the equation (i). Therefore, we have
 $\begin{align}
  & {{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4} \\
 & \Rightarrow {{\tan }^{-1}}\left( 2\left( -1 \right) \right)+{{\tan }^{-1}}\left( 3\left( -1 \right) \right)=\dfrac{\pi }{4} \\
 & {{\tan }^{-1}}\left( -2 \right)+{{\tan }^{-1}}\left( -3 \right)=\dfrac{\pi }{4} \\
\end{align}$
The term ${{\tan }^{-1}}\left( -x \right)$ can be written as $-{{\tan }^{-1}}\left( x \right)$ therefore, the equation gets converted into $-{{\tan }^{-1}}\left( 2 \right)-{{\tan }^{-1}}\left( 3 \right)=\dfrac{\pi }{4}$. As we can clearly see that although the value of x = - 1 is resulting into the R.H.S. of the equation but, the left side of the equation becomes negative which is not acceptable. Therefore x = -1 does not satisfy this expression (i).
Now, we will substitute the value of $x=\dfrac{1}{6}$ in the equation (i). Thus we get,
${{\tan }^{-1}}\left( 2\times \dfrac{1}{6} \right)+{{\tan }^{-1}}\left( 3\times \dfrac{1}{6} \right)=\dfrac{\pi }{4}$. Thus, we get that ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{4}$. Since, the left side is positive therefore, x = $\dfrac{1}{6}$ is the right answer.
Hence, x = $\dfrac{1}{6}$ is the solution of ${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4}$.

Note: We could have solved the equation $6{{x}^{2}}+5x-1=0$ by hit and trial method. In this method we actually substitute the values as x = 0, 1, -1, and so on. And we collect the first number that satisfies the quadratic equation. After that we divide the equation by that factor and get the desired values of the equation. Due to the two values of x raised the need of cross checking the values and because of this one value is always rejected while the other value gets into the solution.