Question

Show that :
$\tan {48}^o\tan {23}^o\tan {42}^o\tan {67}^o=1$

Answer 1

$\tan {48}^0\tan {23}^0\tan {42}^0\tan {67}^0$

$=\tan ({90}^0-{42}^0)\tan ({90}^0-{67}^0)\tan {42}^0\tan {67}^0$

$=\cot {42}^0\cot {67}^0\tan {42}^0\tan {67}^0$

$=\frac{1}{\tan {42}^0}\frac{1}{\tan {67}^0}\tan {42}^0\tan {67}^0$

$=1$