Question

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

(a) $\frac{3-\sqrt{5}}{2}$

(b) $\frac{3+\sqrt{5}}{2}$

(c) $\frac{2}{3-\sqrt{5}}$

(d) $\frac{2}{3+\sqrt{5}}$

• A. $aandd$
• B. $donly$
• C. $bonly$
• D. $Noneofthese$

Answer 1

The correct option is A

$aandd$

Explanation for the correct option:

Finding the value of given equation:

Given , $\tan \left[\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]$

Let us assume, $\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)=\theta$

By multiplying both sides by $2$

$$\begin{gathered} \Rightarrow \\ \Rightarrow \end{gathered}$$ $$\begin{gathered} 2\times \left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)=2\times \theta \\ {\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)=2\theta \end{gathered}$$

and taking $\cos$on both side

$$\begin{gathered} \Rightarrow \cos 2\theta =\cos \left({\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \\ \Rightarrow \cos 2\theta =\frac{\sqrt{5}}{3} \end{gathered}$$

Using Trigonometric Identity $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$ we get,

$$\begin{gathered} \Rightarrow \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{\sqrt{5}}{3} \\ \Rightarrow 3\left(1-{\tan }^2\theta \right)=\sqrt{5}\left(1+{\tan }^2\theta \right) \\ \Rightarrow 3-3{\tan }^2\theta =\sqrt{5}+\sqrt{5}{\tan }^2\theta \\ \Rightarrow 3-\sqrt{5}=\sqrt{5}{\tan }^2\theta +3{\tan }^2\theta \\ \Rightarrow 3-\sqrt{5}=\left(3+\sqrt{5}\right){\tan }^2\theta \\ \Rightarrow {\tan }^2\theta =\frac{3-\sqrt{5}}{3+\sqrt{5}}\to \left(i\right) \end{gathered}$$

On multiplying and dividing by $3-\sqrt{5}$ in equation $\left(i\right)$

$$\begin{gathered} \Rightarrow {\tan }^2\theta =\frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}} \\ \Rightarrow {\tan }^2\theta =\frac{{(3-\sqrt{5})}^2}{4} \\ \Rightarrow \tan \theta =\frac{(3-\sqrt{5})}{2} \end{gathered}$$

Also multiplying and dividing by $3+\sqrt{5}$ in equation $\left(i\right)$

$$\begin{gathered} \Rightarrow {\tan }^2\theta =\frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}} \\ \Rightarrow {\tan }^2\theta =\frac{4}{{\left(3+\sqrt{5}\right)}^2} \\ \Rightarrow \tan \theta =\frac{2}{\left(3+\sqrt{5}\right)} \end{gathered}$$

Therefore, we get two values as

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

$t\mathrm{an}\left[\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]=\frac{2}{3+\sqrt{5}}\left[\text{where}\theta =\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]$

Hence, option (A) is the correct answer.