If tan theta = frac{{x,sin ,phi }}{{1 - x,cos | vedclass

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Question

(b) We have $	an 	heta = \frac{{x\,\sin \,\phi }}{{1 - x\,\cos \,\,\phi }}$

$ \Rightarrow \,\,\frac{1}{x}	an 	heta - 	an 	heta \,\,

Vedclass · Accepted Answer

$\frac{{\sin 	heta }}{{\sin \phi }}$

Vedclass · Answer

(b) We have $	an 	heta = \frac{{x\,\sin \,\phi }}{{1 - x\,\cos \,\,\phi }}$

$ \Rightarrow \,\,\frac{1}{x}	an 	heta - 	an 	heta \,\,\cos \phi = \sin \,\phi $ 

$ \Rightarrow \,\,\frac{1}{x} = \frac{{\sin \,\phi + \cos \,\,\phi \,	an \,	heta }}{{	an \,	heta }}$

and $	an \,\phi = \frac{{y\,\sin \,	heta }}{{1 - y\,\cos \,	heta }}$

$ \Rightarrow 	an \,\phi \, = \frac{{\sin \,	heta }}{{\frac{1}{y} - \cos \,	heta }}$

$ \Rightarrow \,\,\frac{1}{y}	an \,\phi - 	an \,\phi \cos 	heta = \sin 	heta $

$ \Rightarrow \,\,\frac{1}{y}	an \,\phi \, = \sin \,	heta + 	an \,\phi \,\cos 	heta $

$	herefore \,\,\,\,\frac{1}{y} = \frac{{\sin \,	heta + 	an \,\phi \,\cos 	heta }}{{	an \,\phi }}$

Now $\frac{x}{y} = \left[ {\frac{{	an \,	heta }}{{\sin \,\phi + \cos \,\phi \,	an \,	heta }}} ight] 	imes \left[ {\frac{{\sin \,	heta + 	an \,\phi \,\cos \,	heta }}{{	an \,\phi }}} ight]$

$ = \frac{{	an \,	heta }}{{	an \,\phi }}\,\left[ {\frac{{\sin \,	heta + \cos \,	heta \frac{{\sin \varphi }}{{\cos \phi }}}}{{\sin \phi + \cos \phi \frac{{\sin 	heta }}{{\cos 	heta }}}}} ight] $

$= \frac{{	an 	heta \,\,\cos 	heta }}{{	an \,\phi \,\cos \phi }} = \frac{{\sin 	heta }}{{\sin \phi }}$

Aliter : $x\,\sin \,\phi = 	an \,	heta - x\,\cos \,\phi \,	an \,	heta $

$ \Rightarrow \,x = \frac{{	an \,	heta }}{{\sin \,\phi + \cos \,\phi \,	an \,	heta }}$

$ = \frac{{\sin \,	heta }}{{\cos \,	heta \sin \,\phi + \cos \,\phi \,\sin \,	heta }} = \frac{{\sin \,	heta }}{{\sin \,(	heta + \phi )}}$

Similarly, $y = \frac{{\sin \,\phi }}{{\sin \,(	heta + \phi )}}$;

$	herefore \,\,\frac{x}{y} = \frac{{\sin 	heta }}{{\sin \phi }}.$

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

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