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सिद्ध कीजिए कि - `tan 6^(@)tan 42^(@) tan 66^(@) tan 78^(@)=1` |
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Answer» बायाँ पक्ष `=tan6^(@)tan42^(@)tan66^(@)tan78^(@)` `=(sin 6^(@))/(cos 6^(@)).(sin 42^(@))/(cos 42^(@)).(sin66^(@))/(cos 66^(@)).(sin 78^(@))/(cos 78^(@))` `=(2sin 66^(@) sin 6^(@))/(2cos 6^(@) cos 6^(@)).(2sin 78^(@) sin 42^(@))/(2cos 78^(@) cos 42^(@))` `=(cos 60^(@)-cos 72^(@))/(cos 60^(@)+cos 72^(@)).(cos 36^(@)-cos 120^(@))/(cos 36^(@)+cos 120^(@))` `=(((1)/(2)-sin 18^(@))(cos 36^(@)+(1)/(2)))/(((1)/(2)+sin 18^(@))(cos 36^(@)-(1)/(2)))=((1-2sin 18^(@))/(1+2sin 18^(@)))((2cos 36^(@)+1)/(2cos 36^(@)-1))` `=({1-2((sqrt5-1)/(2))}.{2((sqrt5+1)/(4))+1})/({1+2((sqrt5-1)/(4))}.{2((sqrt5+1)/(4))-1})` `=(2-sqrt5+1)/(2+sqrt5-1)xx(2sqrt5+2+4)/(2sqrt5+2-4)` `=(3-sqrt5)/(1+sqrt5)xx(3+sqrt5)/(sqrt5-1)=(9-5)/(5-1)=1=` दायाँ पक्ष |
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