Question

${\tan }^{-1}\left(\frac{1}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{4x+1}\right)={\tan }^{-1}\frac{2}{x^2}$ has

• A. One Solution
• B. Two Solutions
• C. Three Solutions
• D. No Solution

Answer 1

The correct option is B

Two Solutions

Explanation for the correct option:

Step 1: Finding the value

Given equation: ${\tan }^{-1}\left(\frac{1}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{4x+1}\right)={\tan }^{-1}\frac{2}{x^2}$

Applying Trigonometric Identity ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$ in the given equation, we get

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{2x+1}+{\displaystyle \frac{1}{4x+1}}}{1-\left({\displaystyle \frac{1}{2x+1}}\right)\left({\displaystyle \frac{1}{4x+1}}\right)}\right)={\tan }^{-1}\frac{2}{x^2}$

On Solving the above equation for $x$ we get,

$\Rightarrow {\tan }^{-1}\frac{6x+2}{8x^2+6x}={\tan }^{-1}\frac{2}{x^2}$

$\Rightarrow \tan \left({\tan }^{-1}\frac{6x+2}{8x^2+6x}\right)=\frac{2}{x^2}$

$\Rightarrow \frac{3x+1}{8x+6}=\frac{1}{x}$

$\Rightarrow 3x^2-7x-6=0$

Step 2: Now, Solve the above Quadratic Equation

solving the above quadratic equation, we get

$x=\frac{7\pm \sqrt{49+72}}{6}$

$\Rightarrow x=\frac{7\pm 11}{6}$

$\Rightarrow x=3$ and $x=-\frac{2}{3}$

Therefore, Only Two Values of $x$ are Possible

Hence, option (B) is the correct answer.