Question

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

Answer 1

$\tan ({90}^o-{48}^o)\tan ({90}^o-{23}^o)\tan {42}^o\tan {67}^o=1$

$tan({90}^o-\theta )=cot\theta$

$\cot {42}^o\cot {67}^o\tan {42}^o\tan {67}^o=1$

$\cot {42}^o\tan {42}^o\cot {67}^o\tan {67}^o=1$

$tan\theta \times \cot \theta =1$

$1\times 1\times 1\times 1=1$