C-09-Q109mediumsingle_mcqIf cotA=ba\cot A = \dfrac{b}{a}cotA=ab, then asinA−bcosAasinA+bcosA=?\dfrac{a\sin A - b\cos A}{a\sin A + b\cos A} = ?asinA+bcosAasinA−bcosA=aa2−b2a2+b2\dfrac{a^2 - b^2}{a^2 + b^2}a2+b2a