Question

If $sin{17}^{\circ }=\frac{x}{y}$, then the value of $sec{17}^{\circ }-sin{73}^{\circ }$ will be

• A. $\frac{y^2}{x\sqrt{y^2-x^2}}$
• B. $\frac{x^2}{y\sqrt{y^2-x^2}}$
• C. $\frac{x^2}{y\sqrt{x^2-y^2}}$
• D. $\frac{y^2}{x\sqrt{x^2-y^2}}$

Answer 1

The correct option is B

$\frac{x^2}{y\sqrt{y^2-x^2}}$

We know ${73}^{\circ }+{17}^{\circ }={90}^{\circ }$

$\therefore sec{17}^{\circ }=\frac{y}{\sqrt{y^2-x^2}}$

$sin{73}^{\circ }=\frac{\sqrt{y^2-x^2}}{y}$

$sec{17}^{\circ }-sin{73}^{\circ }$ $=\frac{y}{\sqrt{y^2-x^2}}-\frac{\sqrt{y^2-x^2}}{y}=\frac{y^2-(y^2-x^2)}{y\sqrt{y^2-x^2}}=\frac{y^2-y^2+x^2}{y\sqrt{y^2-x^2}}=\frac{x^2}{y\sqrt{y^2-x^2}}$