If {left( {frac{{sin theta }}{{sin phi }}} ri | vedclass

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Question

(A) Given: ${\left( {\frac{{\sin 	heta }}{{\sin \phi }}} ight)^2} = \frac{{	an 	heta }}{{	an \phi }} = 3$ From the equality ${\left( {\frac{{\si

Vedclass · Accepted Answer

$	heta = n\pi \pm \frac{\pi }{3},\,\phi = n\pi \pm \frac{\pi }{6}$

Vedclass · Answer

(A) Given: ${\left( {\frac{{\sin 	heta }}{{\sin \phi }}} ight)^2} = \frac{{	an 	heta }}{{	an \phi }} = 3$ From the equality ${\left( {\frac{{\sin 	heta }}{{\sin \phi }}} ight)^2} = \frac{{\sin 	heta \cos \phi }}{{\cos 	heta \sin \phi }}$,we get: $\frac{{\sin^2 	heta }}{{\sin^2 \phi }} = \frac{{\sin 	heta \cos \phi }}{{\cos 	heta \sin \phi }}$ $\Rightarrow \sin 	heta \cos 	heta = \sin \phi \cos \phi $ $\Rightarrow \sin 2	heta = \sin 2\phi $ This implies $2	heta = n\pi + (-1)^n 2\phi$. Also,we are given $\frac{{	an 	heta }}{{	an \phi }} = 3$. Since $\frac{{\sin^2 	heta }}{{\sin^2 \phi }} = 3$,we have $\sin^2 	heta = 3 \sin^2 \phi$. Using $\frac{{	an 	heta }}{{	an \phi }} = 3$,we have $	an 	heta = 3 	an \phi$. Substituting $	an \phi = \frac{{	an 	heta }}{3}$ into the identity $\sin^2 	heta = 3 \sin^2 \phi$,we find $	an^2 	heta = 3$,which gives $	heta = n\pi \pm \frac{\pi }{3}$. Then $	an \phi = \frac{{	an 	heta }}{3} = \pm \frac{{\sqrt{3}}}{3} = \pm \frac{1}{{\sqrt{3}}}$,which gives $\phi = n\pi \pm \frac{\pi }{6}$.

If ${\left( {\frac{{\sin \theta }}{{\sin \phi }}} \right)^2} = \frac{{\tan \theta }}{{\tan \phi }} = 3,$ then the values of $\theta$ and $\phi$ are:

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