Quantitative Aptitude Ques 1366

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

Options:

A) $ \frac{y^{2}-x^{2}}{xy} $

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

C) $ \frac{x^{2}}{y\sqrt{y^{2}+x^{2}}} $

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

Show Answer

Answer:

Correct Answer: D

Solution:

  • Given, $ \sin 17{}^\circ =\frac{x}{y} $
    $ \Rightarrow $ $ \sec 17{}^\circ -\sin 73{}^\circ $ $ =\sec 17{}^\circ -\sin (90{}^\circ -17{}^\circ ) $ $ =\sec 17{}^\circ -\cos 17{}^\circ $ $ [\because \sin (90{}^\circ -\theta )=cos] $ $ =\frac{1}{\cos 17{}^\circ }-\cos 17{}^\circ $ $ =\frac{1-{{\cos }^{2}}17{}^\circ }{\cos 17{}^\circ }=\frac{{{\sin }^{2}}17{}^\circ }{\cos 17{}^\circ } $ $ =\frac{\frac{x^{2}}{y^{2}}}{\sqrt{1-\frac{x^{2}}{y^{2}}}} $ $ [\because \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }] $ $ =\frac{\frac{x^{2}}{y^{2}}}{\sqrt{\frac{y^{2}-x^{2}}{y}}}=\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}} $
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