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

For adding inverse tangent functions:

$\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$ when $ab < 1$

When $ab > 1$ and both $a$ and $b$ are positive, add $\pi$ to the result.

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Given: $a = 2$ and $b = 3$

Calculate $ab$:

$ab = 2 \times 3 = 6$

Since $6 > 1$ and both numbers are positive, the second case applies.

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Calculate the fraction:

$\frac{a + b}{1 - ab} = \frac{2 + 3}{1 - 6}$

$= \frac{5}{-5}$

$= -1$

Therefore:

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

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Since $\tan\left(\frac{\pi}{4}\right) = 1$, we have $\tan\left(-\frac{\pi}{4}\right) = -1$

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

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

$= -\frac{\pi}{4} + \frac{4\pi}{4}$

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

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