LHS =asinθ−bcosθasinθ+bcosθ
(divide by cosθ)
=atanθ−batanθ+b
=a×ab−ba×ab+b(tanθ=ab)
=a2−b2a2+b2=RHS
If tanθ=ab, show that asinθ−bcosθasinθ+bcosθ=a2−b2a2+b2
If tanθ=ab, then the value of E =asinθ−bcosθasinθ+bcosθ is−
If acosθ−bsinθ=c, prove that asinθ+bcosθ=±√a2+b2−c2