Question : If $\sin17°=\frac{x}{y}$, then the value of $(\sec17°–\sin73°)$ is:
Option 1: $\frac{y^{2}}{x\sqrt{y^{2}–x^{2}}}$
Option 2: $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$
Option 3: $\frac{x^{2}}{y\sqrt{x^{2}–y^{2}}}$
Option 4: $\frac{y^{2}}{x\sqrt{x^{2}–y^{2}}}$
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Correct Answer: $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$
Solution :
Given: $\sin17°=\frac{x}{y}$
BC = $\sqrt{y^{2}-x^{2}}$
So, $\sec17°=\frac{y}{\sqrt{y^{2}–x^{2}}}$ and $\sin73°=\frac{\sqrt{y^{2}–x^{2}}}{y}$
$\therefore(\sec17°–\sin73°)=\frac{y}{\sqrt{y^{2}–x^{2}}}–\frac{\sqrt{y^{2}–x^{2}}}{y}$= $\frac{y^{2}–(y^{2}–x^{2})}{y\sqrt{y^{2}–x^{2}}}$= $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$
Hence, the correct answer is $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$.
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