| If \[\sin 17{}^\circ =\frac{x}{y},\]then the value of \[\sec 17{}^\circ -\sin 73{}^\circ \]is [SSC (CPO) 2013] |
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) \[\frac{{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}\]
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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