If sin-n sin -2 - then the value of tan - is

The correct option is C 1+n1 ntanαGiven, sinθ =nsin(θ+2α) ⇒1n=sin(θ+2α)sinθ ⇒1+n1 n=sin(θ+2α)+sinθsin(θ+2α) sinθ ⇒1+n1 n=2sin(θ +α)cosα2cos(θ+α)sinα ⇒1+n1 n=ta ...

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Byju's Answer

Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

If sinθ=n sin...

Question

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

\frac{1 - n}{1 + n} cot α

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\frac{1 - n}{1 + n} tan α

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\frac{1 + n}{1 - n} tan α

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\frac{1 + n}{1 - n} cot α

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Solution

The correct option is C $\frac{1 + n}{1 - n} tan α$
Given,
$sin θ = n sin (θ + 2 α)$
$\Rightarrow \frac{1}{n} = \frac{sin (θ + 2 α)}{sin θ}$
$\Rightarrow \frac{1 + n}{1 - n} = \frac{sin (θ + 2 α) + sin θ}{sin (θ + 2 α) - sin θ}$
$\Rightarrow \frac{1 + n}{1 - n} = \frac{2 sin (θ + α) cos α}{2 cos (θ + α) sin α}$
$\Rightarrow \frac{1 + n}{1 - n} = tan (θ + α) cot α$
$\Rightarrow tan (θ + α) = \frac{1 + n}{1 - n} tan α$

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If sinθ=nsin(θ+2α), then the value of tan(θ+α) is

The correct option is C 1+n1−ntanαGiven, sinθ=nsin(θ+2α) ⇒1n=sin(θ+2α)sinθ ⇒1+n1−n=sin(θ+2α)+sinθsin(θ+2α)−sinθ ⇒1+n1−n=2sin(θ+α)cosα2cos(θ+α)sinα ⇒1+n1−n=tan(θ+α)cotα ⇒tan(θ+α)=1+n1−ntanα

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