The value of $\tan^{-1}(2) + \tan^{-1}(3)$ is equal to

$\frac{3\pi}{4}$

The correct answer is Option (3) → $\frac{3\pi}{4}$

$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}\left(\frac{2+3}{1-2\times3}\right)=\tan^{-1}\left(\frac{5}{-5}\right)=\tan^{-1}(-1)$

$\tan^{-1}(-1)=-\frac{\pi}{4}$

But since both angles $\tan^{-1}(2)$ and $\tan^{-1}(3)$ are in the first quadrant, their sum lies in the second quadrant, where $\tan$ is negative.

Hence, $\tan^{-1}(2)+\tan^{-1}(3)=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$

Answer: $\frac{3\pi}{4}$