A) \[\frac{y}{x\sqrt{{{y}^{2}}-{{x}^{2}}}}\]
B) \[\frac{{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}\]
C) \[\frac{{{x}^{2}}}{y\sqrt{{{x}^{2}}-{{y}^{2}}}}\]
D) \[\frac{{{y}^{2}}}{x\sqrt{{{x}^{2}}-{{y}^{2}}}}\]
Correct Answer: B
Solution :
(b) \[\sin {{17}^{{}^\circ }}=\frac{x}{y}\] \[cos{{17}^{{}^\circ }}=\sqrt{1-{{\sin }^{2}}{{17}^{{}^\circ }}}\] \[=\sqrt{1-\frac{{{x}^{2}}}{{{y}^{2}}}}=\sqrt{\frac{{{y}^{2}}-{{x}^{2}}}{{{y}^{2}}}}\] \[\sqrt{\frac{{{y}^{2}}-{{x}^{2}}}{{{y}^{2}}}}\] \[\therefore \sec {{17}^{{}^\circ }}=\frac{y}{\sqrt{{{y}^{2}}-{{x}^{2}}}}\] \[\sin {{73}^{{}^\circ }}=sin\left( {{90}^{{}^\circ }}-{{17}^{{}^\circ }} \right)=cos{{17}^{{}^\circ }}\] \[\therefore sec{{17}^{{}^\circ }}-sin{{73}^{{}^\circ }}\] \[=\frac{y}{\sqrt{{{y}^{2}}-{{x}^{2}}}}-\frac{\sqrt{{{y}^{2}}-{{x}^{2}}}}{y}\] \[=\frac{{{y}^{2}}-{{y}^{2}}+{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}=\frac{{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}\]
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