1
Question
If mtan(θ−30o)=ntan(θ+120o) then prove that cos2θ=(m+n)2(m−n).
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Solution
Now,
mtan(θ−30o)=ntan(θ+120o)
or, tan(θ−30o)tan(θ+120o)=nmor, sin(θ+120o).cos(θ−30o)sin(θ−30o).cos(θ+120o)=mnor, sin(θ+120o).cos(θ−30o)+sin(θ−30o).cos(θ+120o)sin(θ+120o).cos(θ−30o)−sin(θ−30o).cos(θ+120o)=m+nm−n [ By componendo and dividendo]
or, sin{(θ+120o)+(θ−30o)}sin{(θ+120o)−(θ−30o)}=m+nm−nor, sin(90o+2θ)sin150o=m+nm−nor, sin(90o+2θ)cos60o=m+nm−n
or, 2sin(90o+2θ)=m+nm−n
or, 2sin(90o+2θ)=(m+n)2(m−n)
Now,
mtan(θ−30o)=ntan(θ+120o)
or, tan(θ−30o)tan(θ+120o)=nm
or, sin(θ+120o).cos(θ−30o)sin(θ−30o).cos(θ+120o)=mn
or, sin(θ+120o).cos(θ−30o)+sin(θ−30o).cos(θ+120o)sin(θ+120o).cos(θ−30o)−sin(θ−30o).cos(θ+120o)=m+nm−n [ By componendo and dividendo]
or, sin{(θ+120o)+(θ−30o)}sin{(θ+120o)−(θ−30o)}=m+nm−n
or, sin(90o+2θ)sin150o=m+nm−n
or, sin(90o+2θ)cos60o=m+nm−n
or, 2sin(90o+2θ)=m+nm−n
or, 2sin(90o+2θ)=(m+n)2(m−n)
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