Question

If $\tan \theta =\frac{x\sin \varphi }{1-x\cos \varphi }$, $\tan \varphi =\frac{y\sin \theta }{1-y\cos \theta }$ , then $\frac{x}{y}=$

• A. $\frac{\sin \varphi }{\sin \theta }$
• B. $\frac{\sin \theta }{\sin \varphi }$
• C. $\frac{1-cos\varphi }{1-\cos \theta }$
• D. $\frac{1-cos\theta }{1-cos\varphi }$

Answer 1

The correct option is B $\frac{\sin \theta }{\sin \varphi }$

$tan\theta =\frac{xsin\varphi }{1-xcos\varphi }$
$sin\theta -xsin\theta cos\varphi =xcos\theta sin\varphi$
$x=\frac{sin\theta }{sin(\theta +\varphi )}$
Similarly,$y=\frac{sin\varphi }{sin(\theta +\varphi )}$
$\therefore \frac{x}{y}=\frac{sin\theta }{sin\varphi }$