Q.
If sin3θsin(2θ+α)=cos3θcos(2θ+α) and tan2θ=λtan(3θ+α), then the value of λ is _________.
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answer is 2.
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Detailed Solution
We havesin3θsin(2θ+α)=cos3θcos(2θ+α)=k(let)⇒ sin4θsinθsin(2θ+α)=cos4θcosθcos(2θ+α)=k⇒ cos4θ−sin4θ=k[cosθcos(2θ+α)−sinθsin(2θ+α)]⇒ cos2θ=kcos(3θ+α) …. (1)Also,sin3θcosθsin(2θ+α)cosθ=sinθcos3θsinθcos(2θ+α)=k⇒ sin3θcosθ+sinθcos3θ =k(sin(2θ+α)cosθ+sinθcos(2θ+α))⇒ sinθcosθsin2θ+cos2θ=ksin(3θ+α)⇒ sin2θ=2ksin(3θ+α) …… (2)From (1) and (2), we gettan2θ=2tan(3θ+α)
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