If b = a ( + 2 ), then a + b a - b = - Vidyadip

MCQ

If $b\sin \alpha = a\sin (\alpha + 2\beta ),$ then $\frac{{a + b}}{{a - b}} = $

A
$\frac{{\tan \beta }}{{\tan (\alpha + \beta )}}$
B
$\frac{{\cot \beta }}{{\cot (\alpha - \beta )}}$
✓
$\frac{{ - \cot \beta }}{{\cot (\alpha + \beta )}}$
D
$\frac{{\cot \beta }}{{\cot (\alpha + \beta )}}$

✓

Answer

Correct option: C.

$\frac{{ - \cot \beta }}{{\cot (\alpha + \beta )}}$

c
(c) We have $b\,\sin \,\alpha = a\,\sin \,(\alpha + 2\beta )\, $

$\Rightarrow \,\frac{a}{b} = \frac{{\sin \,\alpha }}{{\sin \,(\alpha + 2\beta )}}$

$ \Rightarrow \,\,\frac{{a + b}}{{a - b}} = \frac{{\sin \,\alpha + \sin \,(\alpha + 2\beta )}}{{\sin \,\alpha - \sin \,(\alpha + 2\beta )}} $

$= \frac{{2\,\sin \,(\alpha + \beta )\,\cos \,\beta }}{{ - 2\,\cos \,(\alpha + \beta )\,\sin \,\beta }}$

$ = - \tan \,(\alpha + \beta )\,\cot \,\beta $

$= - \frac{{\cot \beta }}{{\cot \,(\alpha + \beta )}}$.

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