The expression is given to be:
[tex]\tan \left(\frac{7\pi }{12}\right)[/tex]
Rewrite the expression:
[tex]\frac{7\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}[/tex]
Therefore, we have:
[tex]\tan\left(\frac{7\pi}{12}\right)=\tan\left(\frac{\pi}{4}+\frac{\pi}{3}\right)[/tex]
Recall the summation identity:
[tex]\tan \left(x+y\right)=\frac{\tan \left(x\right)+\tan \left(y\right)}{1-\tan \left(x\right)\tan \left(y\right)}[/tex]
Therefore, we have:
[tex]\tan\left(\frac{\pi}{4}+\frac{\pi}{3}\right)=\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\frac{\pi}{3}\right)}{1-\tan\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{3}\right)}[/tex]
Recall that:
[tex]\begin{gathered} \tan \left(\frac{\pi }{4}\right)=1 \\ \tan \left(\frac{\pi }{3}\right)=\sqrt{3} \end{gathered}[/tex]
Hence, the equation becomes:
[tex]\frac{\tan(\frac{\pi}{4})+\tan(\frac{\pi}{3})}{1-\tan(\pi\/4)\tan(\pi\/3)}=\frac{1+\sqrt{3}}{1-1\cdot\sqrt{3}}[/tex]
Therefore, we can simplify the expression to be:
[tex]-2-\sqrt{3}[/tex]
The THIRD OPTION is correct.
