Question

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

Answer 1

To prove: $\tan {48}^{\circ }\tan {23}^{\circ }\tan {42}^{\circ }\tan {67}^{\circ }=1$

LHS $=\tan {48}^{\circ }\tan {23}^{\circ }\tan {42}^{\circ }\tan {67}^{\circ }$

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

$=\cot {42}^{\circ }\cot {67}^{\circ }\tan {42}^{\circ }\tan {67}^{\circ }$ [$\because \tan ({90}^{\circ }-\theta )=\cot \theta$]

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

$=\left(1\right)\left(1\right)$

$=1$

$=$ RHS

Hence, proved.