NIMCET 2018 Question Paper with solutions

Question : 32 of 120

Marks: +1, -0

If sin θ = 3sin (θ + 2α), then the value of tan (θ + α) + 2 tan α is

Solution:

We have
sin θ = 3 sin (θ + 2α)
⇒

{sin(θ+2α)}/{sinθ}

1/3

Apply componendo and dividendo,
⇒

{sin(θ+2α)+sinθ}/{sin(θ+2α)-sinθ}

{1+3}/{1-3}

⇒

{2sin({θ+2α+θ}/2)cos({θ+2α-θ}/2)}/{2cos({θ+2α+θ}/2)sin({θ+2α-θ}/2)}

4/{-2}

⇒

{2sin(θ+α)cosα}/{2cos(θ+α)sinα}

= - 2
⇒

{tan(θ+α)}/{tanα}

= - 2
⇒ tan (θ + α) = - 2 tan α
⇒ tan (θ + α) + 2 tan α = 0

Go to Question:

Prev Question

Next Question