prove that....plzzzzzzz solve this...it's urgent..
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(sin5x-2sin3x+sinx)/(cos5x-cosx)=tanx
LHS
=[(sin5x-sin3x)-(sin3x-sinx)]/(cos5x-cosx)
=[(2cos4x.sinx)-(2cos2x.sinx)]/(-2sin3x.sin2x)
=2sinx.(cos4x-cos2x)/(-2sin3x.2sinx.cosx)
=(cos4x-cos2x)/(-2sin3x.cosx)
=(-2sin3x.sinx)/(-2sin3x,cosx)
=(sinx)/(cosx)
=tanx , Proved.
SECOND -METHOD:-
LHS.
=(sin5x+sinx-2sin3x)/(cos5x-cosx)
=(2sin3x.cos2x-2sin3x)/(-2sin3x.sin2x)
=2sin3x(cos2x-1)(-2sin3x.sin2x)
=(cos2x-1)/(-sin2x)
=(1–2sin^2x-1)/(-2sinx.cosx)
÷(-2sin^2x)/(-2sinx.cosx)
=sinx/cosx
=tanx , Proved.
LHS
=[(sin5x-sin3x)-(sin3x-sinx)]/(cos5x-cosx)
=[(2cos4x.sinx)-(2cos2x.sinx)]/(-2sin3x.sin2x)
=2sinx.(cos4x-cos2x)/(-2sin3x.2sinx.cosx)
=(cos4x-cos2x)/(-2sin3x.cosx)
=(-2sin3x.sinx)/(-2sin3x,cosx)
=(sinx)/(cosx)
=tanx , Proved.
SECOND -METHOD:-
LHS.
=(sin5x+sinx-2sin3x)/(cos5x-cosx)
=(2sin3x.cos2x-2sin3x)/(-2sin3x.sin2x)
=2sin3x(cos2x-1)(-2sin3x.sin2x)
=(cos2x-1)/(-sin2x)
=(1–2sin^2x-1)/(-2sinx.cosx)
÷(-2sin^2x)/(-2sinx.cosx)
=sinx/cosx
=tanx , Proved.
khushi6662:
thanksssss
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