If $\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\sqrt3-2$, then determine the value of $\theta$.
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There will be infinitely many possible values for $\theta$. Have you been given a specific range for $\theta$? – Fly by Night Aug 27 '13 at 15:23
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HINT:
Applying Componendo and dividendo, $$\frac{\sin\theta}{\cos\theta}=\frac{1+\sqrt3-2}{1-(\sqrt3-2)}$$
$$\implies \tan\theta=\frac{\sqrt3-1}{3-\sqrt3}=\frac1{\sqrt3}$$
Can you take it from here?
lab bhattacharjee
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HINT:
Write $\sin\theta-\cos\theta$ in the form $R\sin(\theta-\alpha)$ for appropriate $R$ and $\alpha$. Then write $\sin\theta+\cos\theta$ in the form $R\cos(\theta-\alpha)$. You'll see that it's the same choice of $R$ and $\alpha$. Then your problem becomes $$\frac{R\sin(\theta-\alpha)}{R\cos(\theta-\alpha)} = \sqrt{3}-2$$ Hopefully, you can see that you have $\tan(\theta-\alpha) = \sqrt{3}-2$. Remember: $\alpha$ we be a known number that you already worked out. (If you don't know how to find $R$ and $\alpha$ then leave a comment below.)
Fly by Night
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@ShivamGautam A basic formula is that $\sin(\theta-\alpha) \equiv \sin\theta\cos\alpha-\sin\alpha\cos\theta$. Let's say you want to write $\sin\theta-\cos\theta$ as $R\sin(\theta-\alpha)$. You want $\sin\theta-\cos\theta \equiv R(\sin\theta\cos\alpha-\sin\alpha\cos\theta)$. For this to be true, you need $R\cos\alpha = 1$ and $R\sin\alpha = 1$. Hence $R=1/\sqrt{2}$ and $\alpha = \pi/4$. For the numerator, do the same and expand $R\cos(\theta-\alpha)$ using the identity $\cos(\theta-\alpha) \equiv \cos\theta\cos\alpha + \sin\theta\sin\alpha$. – Fly by Night Aug 27 '13 at 17:16