STD 11 - 3. trignometrical ratios,functions and identities — JEE STD 11 Science — Question

$ = \frac{{\sin \,\,{{20}^o}\sin \,\,{{40}^o}\sin \,\,{{80}^o}\tan {{60}^o}}}{{\cos \,\,{{20}^o}\cos \,\,{{40}^o}\cos \,\,{{80}^o}}}$

Here ${N^r} = (\sin \,\,{20^o}\sin \,\,{40^o}\sin \,\,{80^o})$

$ = \frac{{\sin \,\,{{20}^o}}}{2}\,(2\,\,\sin \,\,{40^o}\sin \,\,{80^o})$

$ = \frac{{\sin \,\,{{20}^o}}}{2}\,(\cos \,\,{40^o} - \cos \,\,{120^o})$

$ = \frac{1}{2}\sin \,\,{20^o}\,\left( {1 - 2\,\,{{\sin }^2}{{20}^o} + \frac{1}{2}} \right)$

$ = \frac{1}{2}\sin \,\,{20^o}\,\left( {\frac{3}{2} - 2\,\,{{\sin }^2}{{20}^o}} \right) = \frac{{\sin \,{{60}^o}}}{4} = \frac{{\sqrt 3 }}{8}$

Now, we take ${D^r} = \cos {20^o}\cos {40^o}\cos {80^o}$

$ = \frac{{\sin \,\,{2^3}\,{{20}^o}}}{{{2^3}\,\sin \,\,{{20}^o}}} = \frac{{\sin \,\,{{160}^o}}}{{8\,\,\sin \,\,{{20}^o}}} = \frac{{\sin \,\,{{20}^o}}}{{8\,\,\sin \,\,{{20}^o}}} = \frac{1}{8}$

$\therefore $ Hence $\tan \,\,{20^o}\tan \,\,{40^o}\tan \,\,{80^o} = \frac{{\sqrt 3 /8}}{{1/8}}$

Therefore $\tan {20^o}\tan {40^o}\tan {60^o}\tan {80^o} = \sqrt 3 .\sqrt 3 = 3$.