Hi Tutors...hope you are having a great day! I managed to find values for delta but could not show solution to equation.
Regards and thanks.
Trigonometry
tanθ=λ tan(A−θ)tanθ=sin(θ)cos(θ)tan(A−θ)=sin(A−θ)cos(A−θ)sin(θ)cos(θ)=λ (sin(A−θ)cos(A−θ))sin(A−θ)=sin(A)cos(θ)−cos(A)sin(θ)cos(A−θ)=cos(A)cos(θ)+sin(A)sin(θ)sin(θ)cos(θ)=λ (sin(A)cos(θ)−cos(A)sin(θ)cos(A)cos(θ)+sin(A)sin(θ))sin(θ)(cos(A)cos(θ)+sin(A)sin(θ))=λ cos(θ)(sin(A)cos(θ)−cos(A)sin(θ))cos(A)sin(θ)cos(θ)+sin(A)sin2(θ)=λ cos2(θ)sin(A)−λ cos(A)sin(θ)cos(θ)(λ+1)cos(A)sin(θ)cos(θ)=λ cos2(θ)sin(A)−sin(A)sin2(θ)cos2(θ)=1−sin2(θ)(λ+1)cos(A)sin(θ)cos(θ)=λ (1−sin2(θ))sin(A)−sin(A)sin2(θ)(λ+1)cos(A)sin(θ)cos(θ)=λ sin(A)−(λ+1)sin(A)sin2(θ)(λ+1)cos(A)sin(θ)cos(θ)+(λ+1)sin(A)sin2(θ)=λ sin(A)(λ+1)(cos(A)sin(θ)cos(θ)+sin(A)sin2(θ))=λ sin(A)sin(θ)cos(θ)=sin(2θ)2sin2(θ)=1−cos(2θ)2(λ+1)(cos(A)sin(2θ)2+sin(A)(1−cos(2θ)2))=λ sin(A)(λ+1)(cos(A)sin(2θ)+sin(A)(1−cos(2θ)))=2λ sin(A)(λ+1)(cos(A)sin(2θ)−sin(A)cos(2θ)+sin(A))=2λ sin(A)cos(A)sin(2θ)−sin(A)cos(2θ)=sin(2θ−A)(λ+1)(sin(2θ−A)+sin(A))=2λ sin(A)(λ+1)sin(2θ−A)+(λ+1)sin(A)=2λ sin(A)(λ+1)sin(2θ−A)=2λ sin(A)−(λ+1)sin(A)(λ+1)sin(2θ−A)=2λ sin(A)−λ sin(A)−sin(A)(λ+1)sin(2θ−A)=λ sin(A)−sin(A)(λ+1)sin(2θ−A)=(λ−1)sin(A)(λ−1)sin(A)=(λ+1)sin(2θ−A)
Trigonometry
tanθ=2 tan(60∘−θ)|λ=2, A=60∘(2−1)sin60∘=(2+1)sin(2θ−60∘)sin60∘=3sin(2θ−60∘)|sin(60∘)=√32√32=3sin(2θ−60∘)sin(2θ−60∘)=√362θ−60∘=arcsin(√36)+360∘n, n∈Z2θ=60∘+16.7786548810∘+360∘n2θ=76.7786548810∘+360∘nθ=38.3893274405∘+180∘nθ1=38.3893274405∘θ2=38.3893274405∘+180∘θ2=218.389327440∘sin(2θ−60∘)=sin(180∘−(2θ−60∘))=sin(180∘−2θ+60∘)=sin(240∘−2θ)sin(240∘−2θ)=√36240∘−2θ=arcsin(√36)+360∘n, n∈Z2θ=240∘−16.7786548810∘−360∘n2θ=223.221345119∘−360∘nθ=111.610672560∘−180∘nθ3=111.610672560∘θ4=111.610672560∘+180∘θ4=291.610672560∘