Question

Find the value of $\dfrac{x}{y}$, if $\tan \theta =\dfrac{x\sin \phi }{1-x\cos \phi }$ and $\tan \phi =\dfrac{y\sin \theta }{1-y\cos \theta }$.

Answer 1

Hint: Find the value of x in terms of $\theta $ and $\phi $ using the equation $\tan \theta =\dfrac{x\sin \phi }{1-x\cos \phi }$ and similarly find the value of y in terms of $\theta $ and $\phi $ using the equation $\tan \phi =\dfrac{y\sin \theta }{1-y\cos \theta }$. Hence find the ratio $\dfrac{x}{y}$. Alternatively use componendo -dividendo, i.e. if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{{{k}_{1}}a+{{k}_{2}}b}{{{k}_{3}}a+{{k}_{4}}b}=\dfrac{{{k}_{1}}c+{{k}_{2}}d}{{{k}_{3}}c+{{k}_{4}}d}$. Hence express x and y in terms of $\theta $ and $\phi $ and hence find the value of $\dfrac{x}{y}$.

Complete step-by-step solution -
We have $\tan \theta =\dfrac{x\sin \phi }{1-x\cos \phi }$
Multiplying both sides by $1-x\cos \phi $, we get
$\tan \theta -x\tan \theta \cos \phi =x\sin \phi $
Adding $x\tan \theta \cos \phi $ on both sides of the equation, we get
$\tan \theta =x\tan \theta \cos \phi +x\sin \phi $
Taking x common from the terms on RHS, we get
$\tan \theta =x\left( \tan \theta \cos \phi +\sin \phi \right)$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence, we have
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{x}{\cos \theta }\left( \sin \theta \cos \phi +\sin \phi \cos \theta \right)$
We know that $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$
Hence, we have
$\sin \theta \cos \phi +\cos \theta \sin \phi =\sin \left( \theta +\phi \right)$
Hence, we have
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{x}{\cos \theta }\left( \sin \left( \theta +\phi \right) \right)$
Multiplying both sides by $\cos \theta $, we get
$\sin \theta =x\sin \left( \theta +\phi \right)$
Dividing both sides by $\sin \left( \theta +\phi \right)$, we get
$x=\dfrac{\sin \left( \theta \right)}{\sin \left( \theta +\phi \right)}\text{ …………... }\left( i \right)$
Also, we have $\tan \phi =\dfrac{y\sin \theta }{1-y\cos \theta }$
Multiplying both sides by $1-y\cos \theta $, we get
$\tan \phi -y\tan \phi \cos \theta =y\sin \theta $
Adding $y\tan \phi \cos \theta $ on both sides of the equation, we get
$\tan \theta =x\tan \theta \cos \phi +x\sin \phi $
Taking y common from the terms on RHS, we get
$\tan \phi =y\left( \tan \phi \cos \theta +\sin \theta \right)$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence, we have
$\dfrac{\sin \phi }{\cos \phi }=\dfrac{x}{\cos \phi }\left( \sin \phi \cos \theta +\sin \theta \cos \phi \right)$
We know that $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$
Hence, we have
$\sin \phi \cos \theta +\cos \phi \sin \theta =\sin \left( \theta +\phi \right)$
Hence, we have
$\dfrac{\sin \phi }{\cos \phi }=\dfrac{y}{\cos \phi }\left( \sin \left( \theta +\phi \right) \right)$
Multiplying both sides by $\cos \phi $, we get
$\sin \phi =y\sin \left( \theta +\phi \right)$
Dividing both sides by $\sin \left( \theta +\phi \right)$, we get
$y=\dfrac{\sin \phi }{\sin \left( \theta +\phi \right)}\text{............................... }\left( ii \right)$
Dividing equation (i) by equation (ii), we get
$\dfrac{x}{y}=\dfrac{\dfrac{\sin \theta }{\sin \left( \theta +\phi \right)}}{\dfrac{\sin \phi }{\sin \left( \theta +\phi \right)}}=\dfrac{\sin \theta }{\sin \phi }$
Hence, we have
$\dfrac{x}{y}=\dfrac{\sin \theta }{\sin \phi }$, which is the required value of $\dfrac{x}{y}$.

Note: Alternative Solution: Best method:
We have
$\tan \theta =\dfrac{x\sin \phi }{1-x\cos \phi }\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{x\sin \phi }{1-x\cos \phi }$
We know that if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{{{k}_{1}}a+{{k}_{2}}b}{{{k}_{3}}a+{{k}_{4}}b}=\dfrac{{{k}_{1}}c+{{k}_{2}}d}{{{k}_{3}}c+{{k}_{4}}d}$.
Hence, we have
$\dfrac{\sin \theta }{\cos \theta \sin \phi +\sin \theta \cos \phi }=\dfrac{x\sin \phi }{\sin \phi \left( 1-x\cos \phi \right)+\cos \phi \left( x\sin \phi \right)}$
Hence, we have
$\dfrac{x\sin \phi }{\sin \phi }=\dfrac{\sin \theta }{\sin \left( \theta +\phi \right)}$
Hence, we have
$x=\dfrac{\sin \theta }{\sin \left( \theta +\phi \right)}$
Similarly, we have
$y=\dfrac{\sin \phi }{\sin \left( \theta +\phi \right)}$ and hence
$\dfrac{x}{y}=\dfrac{\sin \theta }{\sin \phi }$, which is the same as obtained above.