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The value of the expression $\tan\bigg(\frac{1}{2}\cos^{-1}{\frac{2}{\sqrt 5}}\bigg)$ is:

$\begin{array}{1 1} 2 \sqrt 5 \\ \sqrt 5 -2 \\ \frac{\sqrt{5}-2}{2} \\ 5 - \sqrt {2} \end{array} $

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  • \(tan\large\frac{\theta}{2}=\sqrt{\large\frac{1-cos\theta}{1+cos\theta}}\)
Take $ cos^{-1}\large\frac{2}{\sqrt5}$$=\theta\:\:\Rightarrow\:cos\theta=\large\frac{2}{\sqrt{5}}$
\(\Rightarrow\:\sqrt{\large\frac{1-cos\theta}{1+cos\theta}}=\sqrt{\large\frac{1-\frac{2}{\sqrt5}}{1+\large\frac{2}{\sqrt5}}}=\sqrt{\large\frac{\sqrt5-2}{\sqrt5+2}}\)
\(=\sqrt5-2\) (by rationalising the denominator)
By substituting in the above formula we get
$tan\large\frac{\theta}{2}=\sqrt{\large\frac{1-cos\theta}{1+cos\theta}}$$=\sqrt5-2$
$\Rightarrow\:tan\big(\large\frac{1}{2}$$.cos^{-1}\large\frac{2}{\sqrt5}\big)$$=tan\large\frac{\theta}{2}$$=\sqrt5-2$
answered Feb 18, 2013 by thanvigandhi_1
edited Feb 6, 2014 by rvidyagovindarajan_1
 

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