(2) (a2−b2)sinθ+2abcosθ=(a2+b2) On dividing by cosθ, (a2−b2)tanθ+2ab=(a2+b2)secθ On squaring both sides, (a2−b2)2tan2θ+4a2b2+4ab(a2−b2)tanθ=(a2+b2)2sec2θ ⇒(a2−b2)2tan2θ+4ab(a2−b2)tanθ+4a2b2=(a2+b2)2(1+tan2θ) ⇒(a2+b2)2tan2θ−(a2−b2)2tan2θ4ab(a2−b2)2tanθ+(a2+b2)−4a2b2=0 ⇒tan2θ(a2+b2)2−(a2−b2)2−4ab(a2−b2)tanθ+(a2−b2)2=0 ⇒4a2b2tan2θ−4ab(a2−b2)tanθ+(a2−b2)2=0 ⇒2abtanθ−(a2−b2)2=0 ⇒2abtanθ−(a2−b2)=0 ⇒tanθ=