Question

If sin(alpha+bita)/sin(alpha-bita)=a-b/a-b then prove that atanbita=btanalpha

Answer 1

Appropriate Question :-

$\rm \: \dfrac{sin( \alpha + \beta )}{sin(\alpha - \beta)} = \dfrac{a + b}{a - b}, \: prove \: that \: a \: tan\beta = b \: tan\alpha$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \dfrac{sin( \alpha + \beta )}{sin(\alpha - \beta)} = \dfrac{a + b}{a - b}$

On applying Componendo and Dividendo, we get

$\rm \: \dfrac{sin( \alpha + \beta ) + sin(\alpha - \beta)}{sin(\alpha + \beta) - sin(\alpha - \beta)} = \dfrac{a + b + a - b}{a + b - a + b}$

We know,

$\boxed{ \rm{ \:sin(x + y) + sin(x - y) = 2sinx \: cosy \: }} \\ \\ \boxed{ \rm{ \:sin(x + y) - sin(x - y) = 2cosx \: siny \: }}$

So, using these results, we get

$\rm \: \dfrac{2sin\alpha \: cos\beta}{2cos\alpha \: sin\beta} = \dfrac{2a}{2b}$

$\rm \: \dfrac{sin\alpha \: cos\beta}{cos\alpha \: sin\beta} = \dfrac{a}{b}$

$\rm \: \dfrac{tan\alpha}{tan\beta} = \dfrac{a}{b}$

$\rm\implies \:a \: tan\beta \: = \: b \: tan\alpha$

Hence, Proved

$\rule{190pt}{2pt}$

Remark :-

If a : b :: c : d, then Componendo and Dividendo means

$\boxed{ \rm{ \:\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$