A) \[\frac{y}{\sqrt{{{y}^{2}}-{{x}^{2}}}}\]
B) \[\frac{x}{\sqrt{{{x}^{2}}-{{y}^{2}}}}\]
C) \[\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]
D) \[\frac{x}{\sqrt{{{y}^{2}}-{{x}^{2}}}}\]
Correct Answer: B
Solution :
(b): \[\tan \theta =\frac{x}{y}\Rightarrow \cot \theta =\frac{y}{x}\Rightarrow \cos ec\theta \] \[=\sqrt{1+{{\left( \frac{y}{x} \right)}^{2}}}=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{x}\] \[\therefore \] \[\sin \theta =\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]
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