Question

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

• A. $\frac{10}{17}$
• B. $-1$
• C. $-\frac{7}{17}$
• D. $1$

Answer 1

Answer: (D)

We have, $\tan \left(2{\tan }^{-1}\frac{1}{5}-\frac{\pi }{4}\right)$ $=\tan \left({\tan }^{-1}\frac{5}{12}-{\tan }^{-1}1\right)$ $\left[\because 2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}\right]$ $=\tan \left\{{\tan }^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12}}\right)\right\}=\frac{-7}{17}$ So, the given equation is $17x^2-7x-10=0$ $\Rightarrow$ $(x-1)(17x+10)=0$ $\Rightarrow$ $x=1,\frac{-10}{17}$