If sin A-n sin B- then the value of n-1n-1tanA-B2-

We have sin A=n sin B ⇒n1=sin Asin B ⇒n 1n+1=sin A sin Bsin A+sin B ⇒n 1n+1=2 cosA+B2sinA B22 sinA+B2cosA B2 ⇒n 1n+1=tanA B2A+B2 ⇒n 1n+1 tan(A+B2)=tanA B2 ...

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Standard XIII

Mathematics

Transforming Sum or Difference of Trigonometric Ratios into Product

If sin A=n si...

Question

If $sin A = n sin B, then the value of \frac{n - 1}{n + 1} tan \frac{A + B}{2} =$

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Solution

We have
$sin A = n sin B \Rightarrow \frac{n}{1} = \frac{sin A}{sin B}$

$\Rightarrow \frac{n - 1}{n + 1} = \frac{sin A - sin B}{sin A + sin B}$

$\Rightarrow \frac{n - 1}{n + 1} = \frac{2 cos \frac{A + B}{2} sin \frac{A - B}{2}}{2 sin \frac{A + B}{2} cos \frac{A - B}{2}}$

$\Rightarrow \frac{n - 1}{n + 1} = tan \frac{A - B}{2} cot \frac{A + B}{2}$

$\Rightarrow \frac{n - 1}{n + 1} tan (\frac{A + B}{2}) = tan \frac{A - B}{2}$

Hence the correct answer is Option B

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If sin A=n sin B, then the value of n−1n+1tanA+B2=

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