<s>[INST] If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x  + \cot x$?
 [/INST] We have
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}.$$
So you want to know how much is $\sin x\cos x$.
Squaring the equality you know, we have:
\begin{align*}
(\sin x + \cos x)^2 &=  \left(\frac{\sqrt{3}+1}{2}\right)^2\\
\sin^2 x + 2\sin x\cos x + \cos^2 x & = \frac{4 + 2\sqrt{3}}{4}\\
1 + 2\sin x \cos x &= 1 + \frac{\sqrt{3}}{2}\\
2\sin x \cos x &= \frac{\sqrt{3}}{2}\\
\sin x \cos x &= \frac{\sqrt{3}}{4}\\
\tan x + \cot x = \frac{1}{\sin x\cos x} &= \frac{4\sqrt{3}}{3}.
\end{align*}
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