If sin3⁡θsin⁡(2θ+α)=cos3⁡θcos⁡(2θ+α) and tan⁡2θ=λtan⁡(3θ+α) then the value of λ is

We havesin3⁡θsin⁡(2θ+α)=cos3⁡θcos⁡(2θ+α)=k(let)⇒ sin4⁡θsin⁡θsin⁡(2θ+α)=cos4⁡θcos⁡θcos⁡(2θ+α)=k⇒ cos4⁡θ−sin4⁡θ=k[cos⁡θcos⁡(2θ+α)−sin⁡θsin⁡(2θ+α)]⇒ cos⁡2θ=kcos⁡(3θ+α)-----iAlso,sin3⁡θcos⁡θsin⁡(2θ+α)cos⁡θ=sin⁡θcos3⁡θsin⁡θcos⁡(2θ+α)=k⇒ sin3⁡θcos⁡θ+sin⁡θcos3⁡θ =k(sin⁡(2θ+α)cos⁡θ+sin⁡θcos⁡(2θ+α))⇒ sin⁡θcos⁡θsin2⁡θ+cos2⁡θ=ksin⁡(3θ+α)⇒ sin⁡2θ=2ksin⁡(3θ+α)---iiFrom (1) and (2), we get tan⁡2θ=2tan⁡(3θ+α)