sin7x-sin5x/cos7x+cos5x=tanx
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Answered by
17
LHS sin7x-sin5x/cos7x+cos5x=
2cos(7x+5x/2).sin(7x-5x/2)/2cos(7x+5x/2).cos(7x-5x/2)
i.e; cos(12x/2).sin(2x/2)/cos(12x/2).cos(2x/2)
i.e., cos(6x).sin(x)/cos(6x).cos(x)
i.e.,sinx/cosx
i.e.,tanx=RHS
2cos(7x+5x/2).sin(7x-5x/2)/2cos(7x+5x/2).cos(7x-5x/2)
i.e; cos(12x/2).sin(2x/2)/cos(12x/2).cos(2x/2)
i.e., cos(6x).sin(x)/cos(6x).cos(x)
i.e.,sinx/cosx
i.e.,tanx=RHS
ramya01:
ho / is the hole bracket of cos7x+cos5x?
Yes
Answered by
4
L.H.S=sin7x-sin5x/cos7x+cos5x
={2.cos[(7x +5x)/2].sin[(7x-5x)/2]}/{2.cos[(7x+5x)/2].cos[(7x-5x)/2]
( since,sinC-sinD=2.cos[(C+D)/2].sin[(C-D)/2]
cosC+cosD=2.cos[(C+D)/2].cos[(C-D)/2] )
=(cos3x.sinx)/(cos3x.cosx)
=sinx/cosx
=tanx
=R.H.S
=>(sin7x-sin5x)/(cos7x+cos5x)=tanx
={2.cos[(7x +5x)/2].sin[(7x-5x)/2]}/{2.cos[(7x+5x)/2].cos[(7x-5x)/2]
( since,sinC-sinD=2.cos[(C+D)/2].sin[(C-D)/2]
cosC+cosD=2.cos[(C+D)/2].cos[(C-D)/2] )
=(cos3x.sinx)/(cos3x.cosx)
=sinx/cosx
=tanx
=R.H.S
=>(sin7x-sin5x)/(cos7x+cos5x)=tanx
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