1. ## Tan - equation

Prove :

2. $\tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(\frac{5\pi}{12}\right)$
\begin{aligned}
&= \tan^2\left(\frac{\pi}{12}\right) + \cot^2\left(\frac{\pi}{12}\right) \\
&= \left[\csc \left(\frac{\pi}{6}\right) - \cot \left(\frac{\pi}{6}\right)\right]^2 + \left[\csc \left(\frac{\pi}{6}\right) + \cot \left(\frac{\pi}{6}\right)\right]^2
\end{aligned}

\begin{aligned}
&= \csc^2 \left(\frac{\pi}{6}\right) - 2\csc \left(\frac{\pi}{6}\right)\cot \left(\frac{\pi}{6}\right) + \cot^2 \left(\frac{\pi}{6}\right) \\
&\;\;+ \csc^2 \left(\frac{\pi}{6}\right) + 2\csc \left(\frac{\pi}{6}\right)\cot \left(\frac{\pi}{6}\right) + \cot^2 \left(\frac{\pi}{6}\right) \\
&= 2\csc^2 \left(\frac{\pi}{6}\right) + 2\cot^2 \left(\frac{\pi}{6}\right) \\
\end{aligned}

\begin{aligned}
&= 2\csc^2 \left(\frac{\pi}{6}\right) + 2\left(\csc^2 \left(\frac{\pi}{6}\right) - 1\right) \\
&= 4\csc^2 \left(\frac{\pi}{6}\right) - 2 \\
&= 4(2^2) - 2 \\
&= 14
\end{aligned}

Note: this was not meant to be an efficient way.

01

3. Hello, dhiab!

My method is not very efficient either . . .

. . $\begin{array}{ccc}\sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} \\ \\[-4mm] \cos^2\!\theta &=& \dfrac{1+\cos2\theta}{2} \end{array}$

$\text{Prove: }\;\tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(\frac{5\pi}{12}\right) \:=\:14$

We have: . $\frac{\sin^2(\frac{\pi}{12})}{\cos^2(\frac{\pi}{12 })} + \frac{\sin^2(\frac{5\pi}{12})}{\cos^2(\frac{5\pi}{ 12})}$ . $=\;\; \frac{\dfrac{1-\cos\frac{\pi}{6}}{2}} {\dfrac{1 + \cos\frac{\pi}{6}}{2}} + \frac{\dfrac{1-\cos\frac{5\pi}{6}}{2}} {\dfrac{1 + \cos\frac{5\pi}{6}}{2}}$

. . . . . . $= \;\frac{1-\cos\frac{\pi}{6}}{1 + \cos\frac{\pi}{6}} + \frac{1-\cos\frac{5\pi}{6}}{1 + \cos\frac{5\pi}{6}} \;=\;\frac{1-\frac{\sqrt{3}}{2}} {1 + \frac{\sqrt{3}}{2}} + \frac{1+\frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}$

. . . . . . $=\; \frac{\left(1-\frac{\sqrt{3}}{2}\right)^2 + \left(1 + \frac{\sqrt{3}}{2}\right)^2} {1 - \left(\frac{\sqrt{3}}{2}\right)^2} \;=\; \frac{\left(1 - \sqrt{3} + \frac{3}{4}\right) + \left(1 + \sqrt{3} + \frac{3}{4}\right)}{1 - \frac{3}{4}}$

. . . . . . $=\;\frac{\frac{14}{4}}{\frac{1}{4}} \;=\;\boxed{14}$