Question

$\ \textless \ br /\ \textgreater \ \underline{ \boxed{\mathbb{\red{question}}}} \\ \\ \tan( \alpha ) = \frac{x \sin( \beta ) }{1 - x \cos( \beta ) } \: and \: \\ \tan( \beta ) = \frac{y \sin( \alpha ) }{1 - y \cos( \alpha ) } \\ then \: the \: value \: of \: \frac{x}{y} is \:$
please give the correct answer

spammer do not answer

correct answer mark brainlist and get multiple thanks

All the best​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{tan(\alpha)=\dfrac{x\,sin(\beta)}{1-x\,cos(\beta)}\,\,\,\,\,\,\,\,\,\,...(1)}\\\\\tt{tan(\beta)=\dfrac{y\,sin(\alpha)}{1-y\,cos(\alpha)}\,\,\,\,\,\,\,\,\,\,...(2)}$

From (1), we get,

$\tt{\dfrac{sin(\alpha)}{cos(\alpha)}=\dfrac{x\,sin(\beta)}{1-x\,cos(\beta)}}$

$\tt{\implies\,sin(\alpha)-x\,sin(\alpha)\,cos(\beta)=x\,sin(\beta)\,cos(\alpha)}$

$\tt{\implies\,sin(\alpha)=x\,sin(\beta)\,cos(\alpha)+x\,sin(\alpha)\,cos(\beta)}$

$\tt{\implies\,sin(\alpha)=x\left\{sin(\beta)\,cos(\alpha)+sin(\alpha)\,cos(\beta)\right\}}$

$\tt{\implies\,sin(\alpha)=x\,sin(\alpha+\beta)}$

$\tt{\implies\,x=\dfrac{sin(\alpha)}{sin(\alpha+\beta)}}$

From (2), we get,

$\tt{\dfrac{sin(\beta)}{cos(\beta)}=\dfrac{y\,sin(\alpha)}{1-y\,cos(\alpha)}}$

$\tt{\implies\,sin(\beta)-y\,cos(\alpha)\,sin(\beta)=y\,cos(\beta)\,sin(\alpha)}$

$\tt{\implies\,sin(\beta)=y\left\{cos(\beta)\,sin(\alpha)+cos(\alpha)\,sin(\beta)\right\}}$

$\tt{\implies\,sin(\beta)=y\,sin(\alpha+\beta)}$

$\tt{\implies\,y=\dfrac{sin(\beta)}{sin(\alpha+\beta)}}$

So,

$\tt{\dfrac{x}{y}=\dfrac{\dfrac{sin(\alpha)}{sin(\alpha+\beta)}}{\dfrac{sin(\beta)}{sin(\alpha+\beta)}}}$

$\tt{\implies\dfrac{x}{y}=\dfrac{sin(\alpha)}{sin(\beta)}}$