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Q.
Number of solutions of the equation $tan^{-1}\left(\frac{1}{2x+1}\right)+tan^{-1}\left(\frac{1}{4x+1}\right) = tan^{-1}\left(\frac{2}{x^{2}}\right)$ is
Inverse Trigonometric Functions
Solution:
We have,
$tan^{-1}\left(\frac{1}{2x+1}\right)+tan^{-1}\left(\frac{1}{4x+1}\right) = tan^{-1}\left(\frac{2}{x^{2}}\right)$
$\Rightarrow tan^{-1}\left\{\frac{\frac{1}{2x+1}+\frac{1}{4x+1}}{1-\left(\frac{1}{2x+1}\right)\left(\frac{1}{4x+1}\right)}\right\} = tan^{-1}\left(\frac{2}{x^{2}}\right)$
$\Rightarrow tan^{-1} \left(\frac{6x +2}{8x^{2}+6x}\right)=tan^{-1}\left(\frac{2}{x^{2}}\right)$
$\Rightarrow tan \left\{tan^{-1}\left(\frac{2\left(3x+1\right)}{2x\left(4x+3\right)}\right)\right\} = \frac{2}{x^{2}}$
$\Rightarrow \frac{3x+1}{x\left(4x+3\right)} = \frac{2}{x^{2}}$
$\Rightarrow 3x^{2} - 7x - 6 = 0$
$\Rightarrow \left(x-3\right)\left(3x+2\right) = 0$
$\Rightarrow x = 3$ or $x = \frac{-2}{3} $
$\therefore $ The given equation has $2$ solutions.