Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given ${\tan }^{-1}\left(2x\right)+{\tan }^{-1}\left(3x\right)=\frac{\pi }{4}$

$\Rightarrow \mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\left(\frac{2x+3x}{1-2x\times 3x}\right)=\frac{\pi }{4}$

Applying tangent on both sides,

$\Rightarrow \frac{5x}{1-6x^2}=1 \left[\because \tan \frac{\mathrm{\pi }}{4}=1\right]$

$\Rightarrow 5x=1-6x^2$

$\Rightarrow 6x^2+5x-1=0$

$\Rightarrow 6x^2+6x-x-1=0$

$\Rightarrow \left(6x-1\right)\left(x+1\right)=0$

$\Rightarrow x=-1, \frac{1}{6}$

But, since $x\ge 0$.

Hence, $x=\frac{1}{6}$

$\therefore A$ has only one element and thus, it is a singleton.