The value of tan^-1(2) + tan^-1(3) is equal to

For adding inverse tangent functions:

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

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

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

Calculate $ab$ab:

$$ab = 2 \times 3 = 6$$ab=2×3=6

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

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Calculate the fraction:

$$\frac{a + b}{1 - ab} = \frac{2 + 3}{1 - 6}$$1−aba+b​=1−62+3​

$$= \frac{5}{-5}$$=−55​

$$= -1$$=−1

Therefore:

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

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

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

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

$$= -\frac{\pi}{4} + \frac{4\pi}{4}$$=−4π​+44π​

$$= \frac{3\pi}{4}$$=43π​

Therefore, $\tan^{-1}(2) + \tan^{-1}(3) = \frac{3\pi}{4}$tan−1(2)+tan−1(3)=43π​