<s>[INST] Proving a trigonometric identity with tangents Prove that: $$\tan^227^\circ +2 \tan27^\circ \tan36^\circ=1$$
any help, I appreciate it.
 [/INST] The solution is nothing more than some computation and observing $2\cdot27 = 90 - 36.$
\begin{align*}
\tan^2(27^\circ) + 2 \tan(27 ^ \circ) \tan (36^\circ) &= \tan^2(27^\circ) + 2 \tan(27 ^ \circ) \cot (54^\circ) \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ)}{ \tan(54^\circ)} \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ)}{\frac{2 \tan{27^\circ}}{1 - \tan^2(27^\circ)}} \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ) (1 - \tan^2(27^\circ))}{2 \tan{27^\circ}} \\[1ex]
&= \tan^2(27^\circ) +   (1 - \tan^2(27^\circ)) \\
&= 1,
\end{align*}
where we have used the following identities:
$$\tan(\theta) = \cot(90^\circ - \theta),$$
$$\tan(2\theta) = \frac{2 \tan{\theta}}{1 - \tan^2(\theta)}.$$
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