Inverse Trigonometric Functions Question 22
Question: $ \tan [ \frac{1}{2}{{\cos }^{-1}}( \frac{\sqrt{5}}{3} ) ]= $
[Roorkee 1986]
Options:
A) $ \frac{3-\sqrt{5}}{2} $
B) $ \frac{3+\sqrt{5}}{2} $
C) $ \frac{2}{3-\sqrt{5}} $
D) $ \frac{2}{3+\sqrt{5}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \tan [ \frac{1}{2}{{\cos }^{-1}}( \frac{\sqrt{5}}{3} ) ] $ 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 } $
Therefore $ \sqrt{5}+\sqrt{5}{{\tan }^{2}}\theta =3-3{{\tan }^{2}}\theta $
Therefore $ (\sqrt{5}+3){{\tan }^{2}}\theta =3-\sqrt{5}\Rightarrow {{\tan }^{2}}\theta =\frac{3-\sqrt{5}}{3+\sqrt{5}} $
Therefore $ {{\tan }^{2}}\theta =\frac{{{(3-\sqrt{5})}^{2}}}{4}\Rightarrow \tan \theta =\frac{3-\sqrt{5}}{2} $ On rationalising
Therefore $ \tan \theta =\frac{3-\sqrt{5}}{2}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}} $ .
BETA
