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Question: Given $180°<{\theta}<360°$and$\frac{1+\tan{\theta}}{1-\tan{\theta}}=7$, compute the value of $\sin{\theta}+\cos{\theta}$

My Attempt: $$\frac{(1+\tan{\theta})\cos{\theta}}{(1-\tan{\theta})\cos{\theta}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}=7\implies\cos{\theta}+\sin{\theta}=7(\cos{\theta}-\sin{\theta})$$

What should I do next?

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I would write $$\sin(x)+\cos(x)=7\cos(x)-7\sin(x)$$ and then $$8\sin(x)=6\cos(x)$$ therefore $$\tan(x)=\frac{3}{4}$$

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Rewrite as $\;\sin\theta+\cos\theta=\cos\theta(1+\tan \theta)$.

Now solve for $\tan\theta$: $\quad\dfrac{1+\tan \theta}{1-\tan \theta}=7 \implies \tan\theta=\dfrac34$.

One also has $\;\cos^2\theta=\dfrac1{1+\tan^2\theta}=\dfrac{16}{25}$, and as $\pi<\theta<2\pi$, with $\tan\theta>0$, we know $\cos\theta<0$, so that $\cos\theta=-\dfrac45$, and finally $$\sin\theta+\cos\theta=-\frac45\,\frac74=-\frac75.$$

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You could have gotten $\tan \theta = 3/4$ directly from the given condition: if $y = \tan \theta$, then $$7 = \frac{1 + \tan \theta}{1 - \tan \theta} = \frac{1 + y}{1 - y},$$ from which cross multiplication yields $$7(1-y) = 1+y.$$

However, this does not immediately tell you how to get $\cos \theta + \sin \theta$. You'd need to do something more; specifically, you could recognize that $\tan \theta = 3/4$ implies that $\theta$ is an angle in a 3-4-5 Pythagorean triangle, where the side opposite to $\theta$ has length $3$ and the side adjacent to $\theta$ has length $4$. However, we are also given that $$\pi < \theta < 2\pi,$$ so $\theta$ must be in quadrant III. It immediately follows that $$\sin \theta = -3/5, \quad \cos \theta = -4/5,$$ and $$\cos \theta + \sin \theta = -\frac{7}{5}.$$

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$\frac{1+\tan \theta}{1-tan \theta}+1=7+1=8$

$\frac{2}{1-tan\theta}=8$

$tan \theta =\frac{3}{4}$

$\ cos \theta=1/\sqrt{(1+tan^2 \theta)}=\frac{4}{5}$

$\sin \theta=(3/4)\times)(4/5)=3/5$

$\sin \theta+\cos theta=4/5+3/5=7/5$

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