Question
Prove that: $\tan\frac{\pi}{12}+\tan\frac{\pi}{6}+\tan\frac{\pi}{12}\tan\frac{\pi}{6}=1$

Answer

We have,
$45^\circ=30^\circ+15^\circ$
$\Rightarrow\tan45^\circ=\tan(30^\circ+15^\circ)$
$\Rightarrow1=\frac{\tan30^\circ+\tan15^\circ}{1-\tan30^\circ\tan15^\circ}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$\Rightarrow 1-\tan30^\circ\tan30^\circ=\tan15^\circ+\tan30^\circ$
$\Rightarrow 1=\tan15^\circ+\tan30^\circ+\tan30^\circ\tan15^\circ$
$\Rightarrow \tan15^\circ+\tan30^\circ+\tan15^\circ\tan30^\circ=1$
Hence proved.

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