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Question

The value of $6 \tan (\cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{2}{3}),$ is

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Solution

We have,
$\begin{aligned} \cos^{- 1} (\frac{4}{5}) = \tan^{- 1} (\frac{3}{4}) \\ ∴ 6 \tan (\cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{2}{3}) \end{aligned}$
$\begin{array}{l} = 6 \tan (\tan^{- 1} \frac{3}{4} + \tan^{- 1} \frac{2}{3}) \\ = 6 \tan \{\tan^{- 1} (\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}})\} \\ = 6 \tan \{\tan^{- 1} (\frac{17}{6})\} = 17 \end{array}$

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