Ex 3.3, 23 - Prove tan 4x = 4 tan x (1 - tan2 x) / 1 - 6tan2x - 2x 3x formula - Proving

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Ex 3.3, 23 Prove that tan 4 = (4 tan (1 tan2 ) )/(1 6 tan2 +tan4 ) Taking L.H.S. tan 4x = (2 tan 2x)/(1 tan2 2x) tan 4x = 2((2 tan )/(1 2 ))/(1 ((2 tan )/(1 2 ))^2 ) = (((4 tan )/(1 2 )))/(1 ((2 tan )^2/(1 2 )^2 ) ) = (((4 tan )/(1 2 )))/(1 ((4 ^2 )/(1 2 )^2 ) ) = (((4 tan )/(1 2 )))/((((1 2 )^2 4 ^2 )/(1 2 )^2 ) ) = (4 tan )/(1 tan2 ) ((1 tan2 )2 )/((1 2 )2 4 tan2 ) = (4 tan )/1 ((1 tan2 ) )/((1 2 )2 4 tan2 ) = (4 tan (1 2 ))/((1 tan2 )2 4 tan 2 ) Using (a b)2 = a2 +b2 2ab = (4 tan ( 1 tan2 ) )/(( (1)2+( 2 )2 2 1 2 ) 4 2 ) = (4 tan ( 1 tan2 ) )/(( (1)2+( 2 )2 2 1 2 ) 4 2 ) = (4 tan ( 1 tan2 ) )/(1 + tan 4 2 tan2 4 tan2 ) = (4 tan ( 1 tan2 ) )/(1 + tan4 6 4 ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved

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