Since x=2>0, y=3>0 & xy=6>1Then Tan-1(x)+Tan-1(y)=π+Tan-1x+y1-xy=π+Tan-151-6=π-π4=3π4

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${Tan}^{- 1} (2) + {Tan}^{- 1} (3) =$

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$\frac{- π}{4}$

$\frac{3 π}{4}$

$\frac{5 π}{4}$

$\frac{π}{4}$

answer is C.

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Detailed Solution

Since $x = 2 > 0, y = 3 > 0 & xy = 6 > 1$

Then ${Tan}^{- 1} (x) + {Tan}^{- 1} (y) = π + {Tan}^{- 1} [\frac{x + y}{1 - xy}] = π + {Tan}^{- 1} [\frac{5}{1 - 6}] = π - \frac{π}{4} = \frac{3 π}{4}$

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