Question

(sin(30°+a)-cos(60°+a))/(sin(30°+a)+cos(60°+a))

Answer 1

sin(30°+a)-cos(60°+a))/(sin(30°+a)+cos(60°+a)=(sin30cosa+cos30°sina)-(cos60°cosa-sin60°sina)(sin30cosa+cos30°sina)+(cos60°cosa-sin60°sina)=(1/2cosa+ frac{sqrt{3}}{2} sina)-(1/2cosa- frac{sqrt{3}}{2} sina)(1/2cosa+ frac{sqrt3}{2} sina)+(1/2cosa- frac{sqrt3}{2} sina)= sqrt{3} sinacosa= sqrt{3}tga