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

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

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Text solutionVerified

Solution :-

t a n^{- 1} (2) + t a n^{- 1} (3)

= t a n^{- 1} (\frac{2 + 3}{1 - 2 \times 3})

= t a n^{- 1} (\frac{5}{1 - 6})

= t a n^{- 1} (\frac{5}{- 5})

= t a n^{- 1} (- 1)

= - t a n^{- 1} (1)

= \frac{- π}{4}

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