If Sin 31° = x/y, the value of Sec 31° - Sin 59° is
x2⁄y√(y2 - x2)
Given that \(sin 31^\circ = \frac{x}{y}\), we want to find the value of \(sec 31^\circ - sin 59^\circ\).
We know that \(sin 59^\circ = cos (90^\circ - 59^\circ) = cos 31^\circ\).
Therefore, the expression becomes \(sec 31^\circ - cos 31^\circ = \frac{1}{cos 31^\circ} - cos 31^\circ\).
Since \(sin^2 \theta + cos^2 \theta = 1\), we have \(cos^2 31^\circ = 1 - sin^2 31^\circ = 1 - (\frac{x}{y})^2 = \frac{y^2 - x^2}{y^2}\)
Thus, \(cos 31^\circ = \frac{\sqrt{y^2 - x^2}}{y}\)
Substituting this into the expression, we get:
\(\frac{y}{\sqrt{y^2 - x^2}} - \frac{\sqrt{y^2 - x^2}}{y} = \frac{y^2 - (y^2 - x^2)}{y\sqrt{y^2 - x^2}} = \frac{x^2}{y\sqrt{y^2 - x^2}}\)
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