Question

A root of the equation $17{{x}^{2}}+17x\tan \left( 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)-\dfrac{\pi }{4} \right)-10=0$ is
(a) $\dfrac{-10}{17}$
(b) -1
(c) $\dfrac{7}{17}$
(d) 1

Answer 1

Hint: We will apply the inverse trigonometric formula which is given by $2{{\tan }^{-1}}\theta ={{\tan }^{-1}}\left( \dfrac{2\theta }{1-{{\theta }^{2}}} \right)$ where $\theta$ is the angle that lies between -1 and 1. We will use this formula to solve the given equation.

Complete step-by-step answer:
Considering the equation given by $17{{x}^{2}}+17x\tan \left( 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)-\dfrac{\pi }{4} \right)-10=0...(i)$
In this expression we can clearly see an expression given by $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$. To solve this expression we will use the formula of inverse trigonometric formula which is given by $2{{\tan }^{-1}}\theta ={{\tan }^{-1}}\left( \dfrac{2\theta }{1-{{\theta }^{2}}} \right)$ where $\theta$ lies between -1 and 1. Therefore, we have
$2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{2\left( \dfrac{1}{5} \right)}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right)...(ii)$
Since, we know that the square of $\dfrac{1}{5}=\dfrac{1}{25}$ and $2\times \dfrac{1}{5}=\dfrac{2}{5}$ therefore, we have
$2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{1-\dfrac{1}{25}} \right)$
By taking lcm in the expression $1-\dfrac{1}{25}$ we have that the lcm(1,25) is 25. Thus, we get $1-\dfrac{1}{25}=\dfrac{25-1}{25}$ . Now, we will substitute this value in equation (ii). Thus, we have
$\begin{aligned} & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{\dfrac{25-1}{25}} \right) \\\ & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{5}}{\dfrac{24}{25}} \right) \\\ \end{aligned}$
At this step we will use the formula given by $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$. Therefore, we have
$\begin{aligned} & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{2}{5}\times \dfrac{25}{24} \right) \\\ & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right) \\\ \end{aligned}$
Now, we will substitute this value in equation (i). Therefore, we have
$\begin{aligned} & 17{{x}^{2}}+17x\tan \left( 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)-\dfrac{\pi }{4} \right)-10=0 \\\ & 17{{x}^{2}}+17x\tan \left( {{\tan }^{-1}}\left( \dfrac{5}{12} \right)-\dfrac{\pi }{4} \right)-10=0...(iii) \\\ \end{aligned}$
Now, as we know that ${{\tan }^{-1}}\left( \dfrac{\pi }{4} \right)=1$. Thus, if we use inverse tan operation on both the sides of this expression we will have ${{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4} \right) \right)={{\tan }^{-1}}\left( 1 \right)$. Now, we will apply the formula given by ${{\tan }^{-1}}\left( \tan \theta \right)=\theta$ . Thus, we have that $\dfrac{\pi }{4}={{\tan }^{-1}}\left( 1 \right)$ . Now, we will substitute the value of $\theta$ in equation (iii) where $\theta =\dfrac{\pi }{4}$. Therefore, we have
$\begin{aligned} & 17{{x}^{2}}+17x\tan \left( {{\tan }^{-1}}\left( \dfrac{5}{12} \right)-\dfrac{\pi }{4} \right)-10=0 \\\ & 17{{x}^{2}}+17x\tan \left( {{\tan }^{-1}}\left( \dfrac{5}{12} \right)-{{\tan }^{-1}}\left( 1 \right) \right)-10=0 \\\ \end{aligned}$
Now, we will apply the formula given by ${{\tan }^{-1}}x\pm {{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x\pm y}{1\mp xy} \right)$. Therefore, we have
$\begin{aligned} & 17{{x}^{2}}+17x\tan \left( {{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}-1}{1+\left( \dfrac{5}{12} \right)\left( 1 \right)} \right) \right)-10=0 \\\ & 17{{x}^{2}}+17x\tan \left( {{\tan }^{-1}}\left( \dfrac{\dfrac{5-12}{12}}{\dfrac{12+5}{12}} \right) \right)-10=0 \\\ & 17{{x}^{2}}+17x\tan \left( {{\tan }^{-1}}\left( \dfrac{-7}{17} \right) \right)-10=0 \\\ \end{aligned}$
Now, we will apply the formula given by $\tan \left( {{\tan }^{-1}}\left( \theta \right) \right)=\theta$. Thus, we have
$\begin{aligned} & 17{{x}^{2}}+17x\dfrac{-7}{17}-10=0 \\\ & \Rightarrow 17{{x}^{2}}-7x-10=0...(iv) \\\ \end{aligned}$
Clearly, x=0 does not satisfy the equation so we will substitute x=1 in equation (iv). Thus we have
$\begin{aligned} & 17{{\left( 1 \right)}^{2}}-7\left( 1 \right)-10=17-7-10 \\\ & \Rightarrow 17-7-10=17-17 \\\ & \Rightarrow 17-17=0 \\\ \end{aligned}$

Since, x=1 satisfies the equation (iv). Thus (x-1) is a factor of (iv). Now we will divide equation (iv) by the factor (x-1) as follows.
$x-1\overset{17x+10}{\overline{\left){\begin{aligned} & 17{{x}^{2}}-7x-10 \\\ & \underline{\pm 17{{x}^{2}}\mp 17x} \\\ & +10x-10 \\\ & \underline{\pm 10x\mp 10} \\\ & \underline{00} \\\ \end{aligned}}\right.}}$
Thus, we have that
$\begin{aligned} & 17{{x}^{2}}-7x-10=\left( x-1 \right)\left( 17x+10 \right) \\\ & \Rightarrow \left( x-1 \right)\left( 17x+10 \right)=0 \\\ \end{aligned}$
Now, we have that x-1=0 or 17x+10=0. We will first consider x-1=0. Therefore we have the value of x=1. Now we will consider the value 17x+10=0. Thus we get 17x=-10 or $x=\dfrac{-10}{17}$.

Note: We could have found the roots of the equation $17{{x}^{2}}-7x-10=0$ alternatively by square root formula. The formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where x are roots of the of the form $a{{x}^{2}}+bx+c=0$. We could have applied the formula $2{{\tan }^{-1}}\left( \theta \right)={{\tan }^{-1}}\left( \dfrac{2\theta }{1-{{\theta }^{2}}} \right)$ into the equation $17{{x}^{2}}+17x\tan \left( 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)-\dfrac{\pi }{4} \right)-10=0$ directly. This resulted in minimum steps.