Question

It ${\tan }^{-1}\frac{x+1}{x-1}+{\tan }^{-1}\frac{x-1}{x}={\tan }^{-1}$ (−7), then the value of x is
(a) 0
(b) −2
(c) 1
(d) 2

Answer 1

(d) 2

We know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$.

$$\begin{gathered} \therefore {\tan }^{-1}\left(\frac{x+1}{x-1}\right)+{\tan }^{-1}\left(\frac{x-1}{x}\right)={\tan }^{-1}\left(-7\right) \\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\frac{x+1}{x-1}\times \frac{x-1}{x}}\right)={\tan }^{-1}\left(-7\right) \\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{x^2+x+x^2-2x+1}{x\left(x-1\right)}}{\frac{x^2-x-x^2+1}{x\left(x-1\right)}}\right)={\tan }^{-1}\left(-7\right) \\ \Rightarrow {\tan }^{-1}\left(\frac{2x^2-x+1}{-x+1}\right)={\tan }^{-1}\left(-7\right) \end{gathered}$$

So, we get

$$\begin{gathered} \frac{2x^2-x+1}{-x+1}=-7 \\ \Rightarrow 2x^2-x+1=7x-7 \\ \Rightarrow 2x^2-8x+8=0 \\ \Rightarrow x^2-4x+4=0 \\ \Rightarrow {\left(x-2\right)}^2=0 \\ \Rightarrow x=2 \end{gathered}$$