Question

If ${\tan }^{-1}\left(\frac{x-1}{x+1}\right)+{\tan }^{-1}\left(\frac{2x-1}{2x+1}\right)={\tan }^{-1}\left(\frac{23}{36}\right)$, then $x=$

• A. $\frac{3}{4},-\frac{3}{8}$
• B. $\frac{3}{4},\frac{3}{8}$
• C. $\frac{4}{3},\frac{3}{8}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Resolve the terms:

$\begin{array}{l}{\tan }^{-1}\left(\frac{x-1}{x+1}\right)+{\tan }^{-1}\left(\frac{2x-1}{2x+1}\right)={\tan }^{-1}\left(\frac{23}{36}\right)\\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{\left(x-1\right)}{\left(x+1\right)}+\frac{\left(2x-1\right)}{\left(2x+1\right)}}{1-\frac{\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(2x+1\right)}}\right)={\tan }^{-1}\left(\frac{23}{36}\right)\mathbf{[}\mathbf{\because }{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{A}\mathbf{+}\mathbf{B}}{\mathbf{1}\mathbf{-}\mathbf{AB}}\mathbf{\right)}\mathbf{]}\\ \Rightarrow \frac{\left(x-1\right)}{\left(x+1\right)}+\frac{\left(2x-1\right)}{\left(2x+1\right)}=\frac{23}{36}\left(1-\frac{\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(2x+1\right)}\right)\\ \Rightarrow 36\left[(2x+1)(x-1)+(2x-1)(x+1)\right]=23\left[(x+1)(2x+1)-(x-1)\left(\right(2x-1)\right]\\ \Rightarrow 24x^2-23x-12=0\\ \Rightarrow \left(3x-4\right)\left(8x+3\right)=0\\ \Rightarrow x=\frac{4}{3},\frac{-3}{8}\end{array}$

Hence, option (D) is correct