A) \[\frac{3-\sqrt{5}}{2}\]
B) \[\frac{3+\sqrt{5}}{2}\]
C) \[\frac{2}{3-\sqrt{5}}\]
D) \[\frac{2}{3+\sqrt{5}}\]
Correct Answer: A
Solution :
\[\tan \left[ \frac{1}{2}{{\cos }^{-1}}\left( \frac{\sqrt{5}}{3} \right) \right]\] Let \[\frac{1}{2}{{\cos }^{-1}}\frac{\sqrt{5}}{3}=\theta \Rightarrow \cos 2\theta =\frac{\sqrt{5}}{3}\] But \[\cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\Rightarrow \frac{\sqrt{5}}{3}=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] Þ\[\sqrt{5}+\sqrt{5}{{\tan }^{2}}\theta =3-3{{\tan }^{2}}\theta \] Þ \[(\sqrt{5}+3){{\tan }^{2}}\theta =3-\sqrt{5}\Rightarrow {{\tan }^{2}}\theta =\frac{3-\sqrt{5}}{3+\sqrt{5}}\] Þ \[{{\tan }^{2}}\theta =\frac{{{(3-\sqrt{5})}^{2}}}{4}\Rightarrow \tan \theta =\frac{3-\sqrt{5}}{2}\] On rationalising Þ \[\tan \theta =\frac{3-\sqrt{5}}{2}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}\].
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