If \( \sin x=\frac{7}{25} \) and \( 90^{\circ}

Ask by Ross Savage. in South Africa

Jan 21,2025

We start with the given information: \( \sin x = \frac{7}{25} \) and \( 90^{\circ} < x < 270^{\circ} \). Since \( x \) is in the second quadrant where sine is positive and cosine is negative, we can find \( \cos x \) using the Pythagorean identity: \[ \sin^2 x + \cos^2 x = 1. \] Plugging in our value for \( \sin x \): \[ \left(\frac{7}{25}\right)^2 + \cos^2 x = 1, \] \[ \frac{49}{625} + \cos^2 x = 1. \] To isolate \( \cos^2 x \), we subtract \( \frac{49}{625} \) from 1: \[ \cos^2 x = 1 - \frac{49}{625} = \frac{625}{625} - \frac{49}{625} = \frac{576}{625}. \] Taking the square root, since \( \cos x \) is negative in the second quadrant: \[ \cos x = -\sqrt{\frac{576}{625}} = -\frac{24}{25}. \] Next, we need to find \( \tan x \): \[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{7}{25}}{-\frac{24}{25}} = -\frac{7}{24}. \] We now substitute these values into the expression \( 25 \cos x - \frac{7}{\tan x} \): \[ 25 \cos x = 25 \left(-\frac{24}{25}\right) = -24. \] Next, we compute \( \frac{7}{\tan x} \): \[ \frac{7}{\tan x} = \frac{7}{-\frac{7}{24}} = -24. \] Now we can put it all together: \[ 25 \cos x - \frac{7}{\tan x} = -24 - (-24) = -24 + 24 = 0. \] Thus, the final value is: \[ \boxed{0}. \]