Question

Show that tan 48º tan 23º tan 42º tan 67º = 1.

Answer 1

Answer:

hence proved ! !!

hope it is clear to you

Answer 2

Proved that tan 48º tan 23º tan 42º tan 67º = 1

Solution:

We need to Show that tan 48º tan 23º tan 42º tan 67º = 1

Let’s solve L.H.S  

$\tan 48^{\circ} \times \tan 42^{\circ} \times \tan 23^{\circ} \times \tan 67^{\circ}$

$\tan 48^{\circ} \times \tan \left(90^{\circ}-48^{\circ}\right) \times \tan 23^{\circ} \times \tan \left(90^{\circ}-23^{\circ}\right)$

We know that $\tan \left(90^{\circ}-x\right)=\cot x$

$\tan 48^{\circ} \times \cot 48^{\circ} \times \tan 23^{\circ} \times \cot 23^{\circ}$

We know that $cot x = \frac{1}{tan x}$

$=\tan 48^{\circ} \times \frac{1}{\tan 48^{\circ}} \times \tan 23^{\circ} \times \frac{1}{\tan 23^{\circ}}$

$= 1 \times 1 = 1$

Hence proved that tan 48º tan 23º tan 42º tan 67º = 1.