If $x = (\tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8})$ , then $\frac{\sin x + \cos x}{\tan x} =$

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$\frac{6 \sqrt{2}}{\sqrt{10}}$

$\frac{12}{\sqrt{10}}$

$\frac{15}{\sqrt{10}}$

$\frac{1}{\sqrt{10}}$

answer is A.

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Detailed Solution

Given that,
$x = \tan^{- 1} (\frac{1}{5}) + \tan^{- 1} (\frac{1}{8})$
As we know, $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$
$∴ x = \tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} \cdot \frac{1}{8}}) = \tan^{- 1} (\frac{\frac{8 + 5}{40}}{\frac{40 - 1}{40}})$
$∴ x = \tan^{- 1} \frac{13}{39} = \tan^{- 1} (\frac{1}{3})$
$∴ \tan x = \frac{1}{3}$
And $\sin x = \frac{1}{\sqrt{10}}$ and $\cos x = \frac{3}{\sqrt{10}}$
Putting values of $\sin x$ , $\cos x$ , $\tan x$
$\frac{\sin x + \cos x}{\tan x} = \frac{\frac{1}{\sqrt{10}} + \frac{3}{\sqrt{10}}}{\frac{1}{3}} = \frac{\frac{4}{\sqrt{10}}}{\frac{1}{3}} = \frac{12}{\sqrt{10}}$