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A SAT II question asks:

If $ \sin\theta + \cos\theta = \dfrac 1 2$, what does $\tan\theta + \cot\theta$ equal?

Which identity would I need to solve this?

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    $\begingroup$ Square the first and simplify the second. $\endgroup$ – Nosrati Jan 19 '17 at 14:49
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Hint

$$\sin\theta+\cos\theta=\frac{1}{2} \implies \left( \sin\theta+\cos\theta \right)^2 = \frac{1}{4} \iff \color{blue}{\cos\theta\sin\theta} = \cdots$$ and $$\tan\theta+\cot\theta = \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} = \frac{\cos^2\theta+\sin^2\theta}{\cos\theta\sin\theta}= \frac{1}{\color{blue}{\cos\theta\sin\theta}} = \cdots$$

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$$\tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\\=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\=\frac1{\sin\theta\cos\theta}\\=\frac1{\frac12+\sin\theta\cos\theta-\frac12}\\=\frac1{\frac12(\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta)-\frac12}\\=\frac1{\frac12(\sin\theta+\cos\theta)^2-\frac12}$$

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Using the fact that the sum of the squares sines and cosines is unity. Square the given expression to obtain the product of sin and cos. Then convert the unknown quantity to the reciprocal of this product.

In particular you should get the product to be $-3/8$ so the answer is $-8/3$.

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    $\begingroup$ @imranfat It was a typo. Sorry about that. $\endgroup$ – Tim The Enchanter Jan 19 '17 at 15:02
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HINT: use that $$\sin(\theta)+\cos(\theta)=\sqrt{2}\sin\left(\theta+\frac \pi 4 \right) = \frac 1 2$$

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