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  4. If tan -1 2 x+ tan -1 3 x=(π/4), then x is equal to

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Q. If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$, then $x$ is equal to

Inverse Trigonometric Functions

Solution:

Given, $ \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}.....$(i)
$ \Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)=\frac{\pi}{4} $
$ \Rightarrow \tan ^{-1}\left(\frac{5 x}{1-6 x^2}\right)=\frac{\pi}{4} $
$ \Rightarrow \frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4} $
$ \Rightarrow \frac{5 x}{1-6 x^2}=1$
$\Rightarrow 5 x=1-6 x^2 $
$ \Rightarrow 6 x^2+5 x-1=0$
$\Rightarrow 6 x^2+6 x-x-1=0$
$\Rightarrow 6 x(x+1)-1(x+1)=0$
$ \Rightarrow (x+1)(6 x-1)=0 \Rightarrow x=\frac{1}{6},-1 $
But $x=-1$ does not satisfy Eq. (i). Since here RHS $>0$ and $\tan ^{-1}(x) < 0$ for $x < 0$.
So, $x=\frac{1}{6}$ is the only solution for given equation.