Problem - 3913
Prove the following identity
\begin{equation}
\tan\alpha + \tan(90^\circ - \alpha)=\frac{2}{\sin 2\alpha}
\end{equation}
By the definition, we have $\tan(90^\circ - \alpha)=\cot\alpha$. Therefore,
\begin{align*}
&\tan\alpha + \tan(90^\circ - \alpha)\\
=\quad&\tan\alpha + \cot\alpha\\
=\quad&\frac{\sin\alpha}{\cos\alpha} + \frac{\cos\alpha}{\sin\alpha}\\
=\quad&\frac{\sin^2\alpha +\cos^2\alpha}{\sin\alpha\cos\alpha}\\
=\quad&\frac{1}{\frac{1}{2}\sin 2\alpha}\\
=\quad&\frac{2}{\sin 2\alpha}
\end{align*}