0
328views
sinx+siny=a and cosx+cosy=b. Prove that tan2(x−y2)=4−a2−b2a2+b2.
1 Answer
written 3.0 years ago by |
Solution:
Given, a=sinx+sinyb=cosx+cosy
∴a2+b2=[sinx+siny]2+[cosx+cosy]2 Formula S+S=2SCC+C=2CC=[2sin(x+y2)cos(x−y2)]2+2cos(x+y2)cos(x−y2)]2=4sin2(x+y2)cos2(x−y2)+4cos2(x+y2)cos2(x−y2)4cos2(x−y2) is common to both areterms=4cos2(x−y2)[sin2(x+y2)+cos2(x+y2)]=4cos2(x−y2)×1∵sin2θ+cos2θ=1a2+b2=4cos2(x−y2)
RHS =4−a2−b2(a2+b2)=4−(a2+b2)(a2+b2)=4−4cos2(x−y2)4cos2(x−y2)=4[1−cos2(x−y2)]4cos2(x−y2)=sin2(x−y2)cos2(x−y2)∵1−cos2θ=sin2θ=tan2(x−y2)= RHS