If sin-sin- - a-ba-b- - a-b- b-0-then find the value of tan -tan -

Given : nbsp; sin(α+β)sin(α β) nbsp;= nbsp;a+ba bApplying nbsp;sin(α+β)=sinαcosβ nbsp;+ nbsp;cosαsinβ, nbsp;we nbsp;get⇒ nbsp;sinαcosβ nbsp; ...

If sinα+βsinα...

Question

$If \frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}, α \neq β; a \neq b; b \neq 0; then find the value of \frac{tan α}{tan β} .$

\frac{a}{b}

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\frac{b}{a}

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\frac{a^{2}}{b^{2}}

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\frac{b^{2}}{a^{2}}

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Solution

The correct option is A $\frac{a}{b}$
Given :
$\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b} Applying sin (α + β) = sin α cos β + cos α sin β, we get \Rightarrow \frac{sin α cos β + cos α sin β}{sin α cos β - cos α sin β} = \frac{a + b}{a - b}$

Applying componendo and dividendo we get:

$\Rightarrow \frac{(sin α cos β + cos α sin β) + (sin α cos β - cos α sin β)}{(sin α cos β + cos α sin β) - (sin α cos β - cos α sin β)} = \frac{(a + b) + (a - b)}{(a + b) - (a - b)}$

$\Rightarrow \frac{2 sin α cos β}{2 cos α sin β} = \frac{2 a}{2 b} \Rightarrow \frac{tan α}{tan β} = \frac{a}{b}$

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