Question

Show that
$tan{48}^{0}tan{23}^{0}tan{42}^{0}tan{67}^0-1$

Answer 1

$\tan {48}^o\tan {23}^o\tan {42}^o\tan {67}^o-1$

$=\tan {48}^o\tan {42}^o\tan {23}^o\tan {67}^o-1$

$=\tan {48}^o\tan (90-48)\tan 23\tan (90-23)-1$

$=\tan 48\cot 48\tan 23\cot 23-1$

$=\left(1\right)\left(1\right)-1=1-1=0$.