Prove that:
$tan^{- 1} a + cot^{- 1} (a + 1) = tan^{- 1} (a^{2} + a + 1)$ .

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Prove that:

tan^{- 1} a + cot^{- 1} (a + 1) = tan^{- 1} (a^{2} + a + 1)

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Text solutionVerified

Since

cot x = \frac{1}{tan x}

\Rightarrow cot^{- 1} x = tan^{- 1} \frac{1}{x}

So, we have,

tan^{- 1} a + cot^{- 1} (a + 1) = tan^{- 1} a + tan^{- 1} \frac{1}{a + 1}

= tan^{- 1} (\frac{a + \frac{1}{a + 1}}{1 - \frac{a}{a + 1}}) = tan^{- 1} (a^{2} + a + 1)

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