If x = tan^{-1} left(frac{1}{5}right) + tan^{ | vedclass

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(A) Given that,$x = 	an^{-1} \left(\frac{1}{5}ight) + 	an^{-1} \left(\frac{1}{8}ight)$.Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1}

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

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(A) Given that,$x = 	an^{-1} \left(\frac{1}{5}ight) + 	an^{-1} \left(\frac{1}{8}ight)$.Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1} \left(\frac{a+b}{1-ab}ight)$,$x = 	an^{-1} \left(\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} \cdot \frac{1}{8}}ight) = 	an^{-1} \left(\frac{\frac{8+5}{40}}{\frac{40-1}{40}}ight) = 	an^{-1} \left(\frac{13}{39}ight) = 	an^{-1} \left(\frac{1}{3}ight)$.Thus,$	an x = \frac{1}{3}$.Since $	an x = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{1}{3}$,the hypotenuse is $\sqrt{1^2 + 3^2} = \sqrt{10}$.Therefore,$\sin x = \frac{1}{\sqrt{10}}$ and $\cos x = \frac{3}{\sqrt{10}}$.Substituting these values into the expression:$\frac{\sin x + \cos x}{	an x} = \frac{\frac{1}{\sqrt{10}} + \frac{3}{\sqrt{10}}}{\frac{1}{3}} = \frac{\frac{4}{\sqrt{10}}}{\frac{1}{3}} = \frac{12}{\sqrt{10}}$.

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

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