If sin A-n sin B- then n-1-n-1tanA-B-2-

We have sin A=n sin B ⇒n/1=sin A/sin B ⇒n 1/n+1=sin A sin B/sin A+sin B=2 cosA+B/2sinA B/2/2 sinA+B/2cosA B/2 =tanA B/2cotA+B/2 ⇒n 1/n+1 tan (A+B/2)=tanA B/2. ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If sin A=n si...

Question

If $s i n A = n s i n B, t h e n \frac{n - 1}{n + 1} t a n \frac{A + B}{2} =$

$s i n \frac{A - B}{2}$

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$t a n \frac{A - B}{2}$

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$c o t \frac{A - B}{2}$

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$N o n e o f t h e s e$

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Solution

The correct option is B
$t a n \frac{A - B}{2}$

We have $s i n A = n s i n B \Rightarrow \frac{n}{1} = \frac{s i n A}{s i n B}$
$\Rightarrow \frac{n - 1}{n + 1} = \frac{s i n A - s i n B}{s i n A + s i n B} = \frac{2 c o s \frac{A + B}{2} s i n \frac{A - B}{2}}{2 s i n \frac{A + B}{2} c o s \frac{A - B}{2}}$
$= t a n \frac{A - B}{2} c o t \frac{A + B}{2}$
$\Rightarrow \frac{n - 1}{n + 1} t a n (\frac{A + B}{2}) = t a n \frac{A - B}{2} .$

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