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Given x = (5)/(13) and 0^ x 90^.

April 24, 2026Mathematics

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Answer

\frac{12}{65}

4 steps

Find $\cos x$ .
Given $\sin x = \frac{5}{13}$ and $0^\circ \le x \le 90^\circ$ .
Using the identity $\sin^2 x + \cos^2 x = 1$ :
$\cos^2 x = 1 - \sin^2 x$ $\cos^2 x = 1 - \left(\frac{5}{13}\right)^2$ $\cos^2 x = 1 - \frac{25}{169}$ $\cos^2 x = \frac{169 - 25}{169}$ $\cos^2 x = \frac{144}{169}$
Since $0^\circ \le x \le 90^\circ$ , $\cos x$ is positive:
$\cos x = \sqrt{\frac{144}{169}} = \frac{12}{13}$
Find $\tan x$ .
Using the identity $\tan x = \frac{\sin x}{\cos x}$ :
$\tan x = \frac{\frac{5}{13}}{\frac{12}{13}}$ $\tan x = \frac{5}{12}$
Substitute the values into the expression $\frac{\cos x - 2 \sin x}{2 \tan x}$ .
Numerator:
$\cos x - 2 \sin x = \frac{12}{13} - 2\left(\frac{5}{13}\right)$ $= \frac{12}{13} - \frac{10}{13}$ $= \frac{2}{13}$
Denominator:
$2 \tan x = 2\left(\frac{5}{12}\right)$ $= \frac{10}{12} = \frac{5}{6}$
Now, divide the numerator by the denominator:
$\frac{\frac{2}{13}}{\frac{5}{6}} = \frac{2}{13} \times \frac{6}{5}$ $= \frac{12}{65}$
The value of the expression is $\boxed{\frac{12}{65}}$ .

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