$tan ({cos}^{- 1} \frac{4}{5} + {tan}^{- 1} \frac{2}{3}) =$ ____

\frac{3}{17}

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\frac{17}{6}

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\frac{17}{4}

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\frac{6}{17}

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Solution

The correct option is B $\frac{17}{6}$
$tan ({cos}^{- 1} \frac{4}{5} + {tan}^{- 1} \frac{2}{3})$
$= tan ({tan}^{- 1} \frac{3}{4} + {tan}^{- 1} \frac{2}{3})$
$= \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}} = \frac{17}{6}$

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