If sinθ+sinΦ=a and cosθ+cosΦ=b then show that bsin(θ+Φ)/2+acos(θ+Φ)/2=0

Steps to solve 
If sinθ+sinΦ=a and cosθ+cosΦ=b then show that bsin(θ+Φ)/2+acos(θ+Φ)/2=0

sinθ+sinΦ=a----(I)

cosθ+cosΦ=b-----(II)

Dividing equation(I)&(II)

⇒sinθ+sinΦ/cosθ+cosΦ=a/b

⇒{2.sin(θ+Φ)/2.cos(θ-Φ)/2}/{2.cos(θ+Φ)/2.cos(θ-Φ)/2}=a/b

⇒{sin(θ+Φ)/2}/{cos(θ+Φ)/2}=a/b

⇒bsin(θ+Φ)/2+acos(θ+Φ)/2=0

Hence if sinθ+sinΦ=a and cosθ+cosΦ=b then bsin(θ+Φ)/2+acos(θ+Φ)/2=0 proved

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