A) \[\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{y}\]
B) \[\frac{\sqrt{{{x}^{2}}-{{y}^{2}}}}{y}\]
C) \[\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]
D) \[\frac{y}{\sqrt{{{x}^{2}}-{{y}^{2}}}}\]
Correct Answer: C
Solution :
\[\tan \,\theta =\frac{x}{y}\] \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \] \[=1+\frac{{{x}^{2}}}{{{y}^{2}}}=\frac{{{y}^{2}}+{{x}^{2}}}{{{y}^{2}}}\] \[\Rightarrow \] \[\sec \theta =\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{y}\] \[\therefore \]\[\cos \theta =\frac{1}{\sec \theta }=\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]
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