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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The numerical value of $\tan 20^{\circ}\tan 80^{\circ}\cot 50^{\circ}$ is equal to

$(a)\;\sqrt 3\qquad(b)\;\large\frac{1}{\sqrt 3}$$\qquad(c)\;2\sqrt 3\qquad(d)\;\large\frac{1}{2\sqrt 3}$

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1 Answer

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$\tan 20^{\circ}\tan 80^{\circ}\cot 50^{\circ}=\tan (50^{\circ}-30^{\circ}).\tan(50^{\circ}+30^{\circ}\large\frac{1}{\tan 50^{\circ}})$
$\qquad\qquad\qquad\qquad\;\;=\large\frac{\tan 50^{\circ}-1/\sqrt 3}{1+\tan 50^{\circ}/\sqrt 3}.\frac{\tan 50^{\circ}+1/\sqrt 3}{1-\tan 50^{\circ}/\sqrt 3}\frac{1}{\tan 50^{\circ}}$
$\qquad\qquad\qquad\qquad\;\;=\large\frac{3\tan^250^{\circ}-1}{3\tan 50^{\circ}-\tan^350^{\circ}}$
$\qquad\qquad\qquad\qquad\;\;=-\cot (150^{\circ})$
$\qquad\qquad\qquad\qquad\;\;=\cot 30^{\circ}$
$\qquad\qquad\qquad\qquad\;\;=\sqrt 3$
Hence (a) is right option.
answered Oct 7, 2013 by sreemathi.v
edited Jan 8, 2014 by sreemathi.v
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