Question

For 20 points
If
\tan( \alpha ) = \frac{x \sin( \beta ) }{1 - x \cos( \beta ) }
​
and

\tan \beta = \frac{y \sin \alpha }{1 - y \cos \alpha }

Then
\frac{x}{y} =

?​​

Answer 1

Answer:

$\frac{x}{y}=\frac{sin\alpha}{sin\beta}$

Step-by-step explanation:

Formula used:

$sin(A+B)=sinA\:cosB+cosA\:sinB$

Given:

$tan\alpha=\frac{x\:sin\beta}{1-x\:cos\beta}$........(1) and

$tan\beta=\frac{y\:sin\alpha}{1-y\:cos\alpha}$ and

$(1)\implies\:tan\alpha-x\:tan\alpha\:cos\beta=x\:sin\beta$

$\implies\:tan\alpha=x\:tan\alpha\:cos\beta+x\:sin\beta$

$\implies\:tan\alpha=x(tan\alpha\:cos\beta+sin\beta)$

$\implies\:tan\alpha=x(\frac{sin\alpha}{cos\alpha}cos\beta+sin\beta)$

$\implies\:tan\alpha=x(\frac{sin\alpha\:cos\beta+cos\alpha\:sin\beta}{cos\alpha})$

$\implies\:\frac{sin\alpha}{cos\alpha}=x(\frac{sin(\alpha+\beta)}{cos\alpha})$

$\implies\:sin\alpha=x\:sin(\alpha+\beta)$

$\implies\:x=\frac{sin\alpha}{sin(\alpha+\beta)}$

similarly

$(2)\implies\:y=\frac{sin\beta}{sin(\alpha+\beta)}$

Now,

$\frac{x}{y}$

$=\frac{\frac{sin\alpha}{sin(\alpha+\beta)}}{\frac{sin\beta}{sin(\alpha+\beta)}}$

$=\frac{sin\alpha}{sin\beta}$

$\implies\:\frac{x}{y}=\frac{sin\alpha}{sin\beta}$

Answer 2

Answer:

what is the question

write it properly or answer will come error