given 3 tan a tan b =1, then prove 2 cos (a+b) =cos (a-b)
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3.tana .tanb = 1
3sina. sinb/cosa.cosb = 1
sina.sinb/cosa.cosb = 1/3
use, componendo dividendo rule ,
(sina.sinb + cosa.cosb)/(cosa.cosb - sina.sinb) = (1 + 3)/(3 -1)
[ use, cosa.cosb+ sina.sinb = cos(a -b)
cosa.cosb - sina.sinb = cos(a + b) ]
cos( a - b)/cos(a + b) = 4/2 = 2
cos(a - b) = 2cos(a + b)
hence proved //
3sina. sinb/cosa.cosb = 1
sina.sinb/cosa.cosb = 1/3
use, componendo dividendo rule ,
(sina.sinb + cosa.cosb)/(cosa.cosb - sina.sinb) = (1 + 3)/(3 -1)
[ use, cosa.cosb+ sina.sinb = cos(a -b)
cosa.cosb - sina.sinb = cos(a + b) ]
cos( a - b)/cos(a + b) = 4/2 = 2
cos(a - b) = 2cos(a + b)
hence proved //
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