$tan^{- 1} (3 / 5) + tan^{- 1} (1 / 4) =$

0

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π / 4

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3 n / 4

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None of these

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Solution

The correct option is B $π / 4$
given, ${tan}^{- 1} \frac{3}{5} + {tan}^{- 1} \frac{1}{4}$

$= {tan}^{- 1} ⎛ ⎝ \frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{5} \times \frac{1}{4}} ⎞ ⎠$

$= {tan}^{- 1} ⎛ ⎝ \frac{\frac{12 + 5}{20}}{1 - \frac{3}{20}} ⎞ ⎠$

$= {tan}^{- 1} ⎛ ⎝ \frac{\frac{17}{20}}{\frac{20 - 3}{20}} ⎞ ⎠$

$= {tan}^{- 1} ⎛ ⎝ \frac{\frac{17}{20}}{\frac{17}{20}} ⎞ ⎠$

$= {tan}^{- 1} (1)$

$= \frac{π}{4}$

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