Solve for x:


Solve for x:

$\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$



To find: value of x

Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$

Given: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$

$\mathrm{LHS}=\tan ^{-1}\left(\frac{2+\mathrm{x}+2-\mathrm{x}}{1-\{(2+\mathrm{x}) \times(2-\mathrm{x})\}}\right)$

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

$=\tan ^{-1} \frac{4}{x^{2}-3}$

Therefore, $\tan ^{-1} \frac{4}{\mathrm{x}^{2}-3}=\tan ^{-1} \frac{2}{3}$

Taking tangent on both sides, we get


$\Rightarrow 12=2 x^{2}-6$

$\Rightarrow 2 x^{2}=18$

$\Rightarrow x^{2}=9$

$\Rightarrow x=3$ or $x=-3$

Therefore, $x=\pm 3$ are the required values of $x$.

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