A) \[\frac{1}{2}\]
B) \[\frac{2}{3}\]
C) \[\frac{1}{3}\]
D) None of these
Correct Answer: C
Solution :
\[3\tan \theta =4\]\[\Rightarrow \]\[\tan \theta =\frac{4}{3}\] \[\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }}=\frac{1-\sin \theta }{\cos \theta }\] \[=\sec \theta -\tan \theta =\sqrt{1+{{\tan }^{2}}\theta }-\tan \theta \] \[=\sqrt{1+{{\left( \frac{4}{3} \right)}^{2}}}-\frac{4}{3}=\sqrt{1+\frac{16}{9}}-\frac{4}{3}=\frac{5}{3}-\frac{4}{3}=\frac{1}{3}\]
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