A) \[\frac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\]
B) \[\frac{b}{a}\]
C) \[\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\]
D) \[\frac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\]
Correct Answer: C
Solution :
| [c] Given, \[\sin \theta =\frac{a}{b}\] |
| \[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\] \[[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\cos \theta =\sqrt{1-{{\left( \frac{a}{b} \right)}^{2}}}=\sqrt{1-\frac{{{a}^{2}}}{{{b}^{2}}}}=\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\] |
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