Question

Show that $tan48tan23tan42tan67$=1

Answer 1

$\tan {48}^{\circ }=\tan ({90}^{\circ }-{42}^{\circ })=\cot {42}^{\circ }=\frac{1}{\tan {42}^{\circ }}$

$\tan {23}^{\circ }=\tan ({90}^{\circ }-{67}^{\circ })=\cot {67}^{\circ }=\frac{1}{\tan {67}^{\circ }}$

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

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

$=1=RHS$

Hence Showed