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

Ask by Wang Byrd. in South Africa

Jan 21,2025

To solve the problem, we start with the given information: \[ \sin x = \frac{7}{25} \] Since \( x \) is in the second quadrant \( (90^{\circ} 0 \) and \( \cos x < 0 \). Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \): \[ \left( \frac{7}{25} \right)^2 + \cos^2 x = 1 \] Calculating \( \left( \frac{7}{25} \right)^2 \): \[ \frac{49}{625} + \cos^2 x = 1 \] Solving for \( \cos^2 x \): \[ \cos^2 x = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625} \] Taking the square root (noting the sign since \( x \) is in the second quadrant): \[ \cos x = -\sqrt{\frac{576}{625}} = -\frac{24}{25} \] Next, we calculate \( \tan x \): \[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{7}{25}}{-\frac{24}{25}} = -\frac{7}{24} \] Now we can find \( \frac{7}{\tan x} \): \[ \frac{7}{\tan x} = \frac{7}{-\frac{7}{24}} = -24 \] Now we can compute \( 25 \cos x - \frac{7}{\tan x} \): \[ 25 \cos x = 25 \left(-\frac{24}{25}\right) = -24 \] Putting it all together: \[ 25 \cos x - \frac{7}{\tan x} = -24 - (-24) = -24 + 24 = 0 \] Therefore, the value of \( 25 \cos x - \frac{7}{\tan x} \) is: \[ \boxed{0} \]