(b)
cos4x−sin4x2sinxcosx=tan2x
taking LHS
=(cos2x+sin2x)(cos2x−sin2x)sin2x=cos2x−sin2xsin2x=cos2xsin2x=tan2x
(a)
cosx+sinxsin2x+cos2x=2cosx
taking LHS
=cosx+sinx2sinxcosx+2cos2x−1
2cosx−sinx+cosx1
so sinx +cosx cannnot be zero so this is not solved further
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