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$$\text{if }\frac{\sin\theta}{\sin\phi}=\frac12 \text{ , }\frac{\cos\theta}{\cos\phi}=\frac32 \text{ ; if both the angles are the acute angle, then find } \tan\theta \text{ and } \tan\phi.$$

this question made me to do all the things i could. I had done multiply, divide, add, subtract those two equation and even the question would between thinking what has been done to him.

but i couldn't solve this question.

Answers: $\frac{\sqrt5}{3\sqrt3}$, $\frac{\sqrt5}{\sqrt3}$ respectively.

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  • $\begingroup$ Your first equation tells you that $\sin \theta = \frac{\sin \phi}{2}$ and your second says $\cos\theta = \frac{3\cos \phi}{2}$. Using the trigonometric pytagorean theorem you'll find $\cos^2 \phi = \frac{3}{8}$. Can you go from here? $\endgroup$ – frog Aug 24 '15 at 13:03
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Hint: We have $$ \sin \phi = 2 \sin \theta\\ \cos \phi = \frac 23 \sin \theta $$ By the Pythagorean identity, we then have $$ \sin^2 \theta + \cos^2 \theta = 1\\ 4\sin^2 \theta + \frac 49 \cos^2 \theta = 1 $$ This is a system of equations that can be solved for $\sin^2 \theta$ and $\cos^2 \theta$.

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dividing, you have $$\tan\theta=\frac 13\tan\phi$$ you also have $$\sec^2\phi=\frac 94\sec^2\theta$$ therefore, $$1+9\tan^2\theta=\frac 94(1+\tan^2\theta)$$ which leads to $$\tan\theta=\frac{\sqrt{5}}{3\sqrt{3}}$$ and so on...

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