Question

$tan\left[\frac{1}{2}co{s}^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]=$

• A. $\frac{3-\sqrt{5}}{2}$
• B. $\frac{3+\sqrt{5}}{2}$
• C. $\frac{2}{3-\sqrt{5}}$
• D. $\frac{2}{3+\sqrt{5}}$

Answer 1

Answer: (A), (D)

$\frac{2}{3+\sqrt{5}}$

$tan\left[\frac{1}{2}co{s}^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]$
Let $\frac{1}{2}co{s}^{-1}\frac{\sqrt{5}}{3}=\theta \Rightarrow cos2\theta =\frac{\sqrt{5}}{3}$
But $cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\Rightarrow \frac{\sqrt{5}}{3}=\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$
$\Rightarrow \sqrt{5}+\sqrt{5}ta{n}^2\theta =3-3ta{n}^2\theta$
$\Rightarrow (\sqrt{5}+3)ta{n}^2\theta =3-\sqrt{5}\Rightarrow ta{n}^2\theta =\frac{3-\sqrt{5}}{3+\sqrt{5}}$
$\Rightarrow ta{n}^2\theta =\frac{(3-\sqrt{5}{)}^2}{4}\Rightarrow tan\theta =\frac{3-\sqrt{5}}{2}$
On rationalising
$\Rightarrow tan\theta =\frac{3-\sqrt{5}}{2}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}$