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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Evaluate : $\tan\big[\large\frac{1}{2}$$\cos^{-1}\large\frac{\sqrt 5}{3}\big]$

$(a)\;\large\frac{3-\sqrt 5}{2}\qquad$$(b)\;\large\frac{3+\sqrt 5}{2}\qquad$$(c)\;0\qquad(d)\;None\;of\;these$

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1 Answer

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Let $\cos^{-1}\big(\large\frac{\sqrt 5}{3}\big)$$=2\theta$
$\cos 2\theta=\large\frac{\sqrt 5}{3}$
$\large\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{\sqrt 5}{3}$
$\large\frac{1+\tan^2\theta}{1-\tan^\theta}=\frac{3}{\sqrt 5}$
$\large\frac{2\tan^2\theta}{2}=\frac{3-\sqrt 5}{3+\sqrt 5}$
$\tan^2\theta=\big(\large\frac{3-\sqrt 5}{3+\sqrt 5}\big)\big(\large\frac{3-\sqrt 5}{3-\sqrt 5}\big)$
$\tan^2\theta=\large\frac{(3-\sqrt 5)^2}{4}$
$\tan\theta=\pm\big(\large\frac{3-\sqrt 5}{2}\big)$------(1)
$\Rightarrow 0\leq 2\theta < \pi$
$\therefore 0\leq \theta\leq \large\frac{\pi}{2}$
$\Rightarrow \theta$ lies in the first quadrant.
Hence $\tan\theta$ must be +ve.
From (1)
$\tan\theta=\large\frac{3-\sqrt 5}{2}$
Hence (a) is the correct option.
answered Oct 17, 2013 by sreemathi.v
 

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