# 2.    Show that : $(i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$

M manish

$\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$
Taking Left Hand Side (LHS)
$\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}$
$\Rightarrow \tan 48^{o}\tan 23^{o}\tan (90^o-48^{o})\tan (90^o-23^{o})$
$\Rightarrow \tan 48^{o}\tan 23^{o}\cot 48^{o}\cot23^{o}$     [it is known that $\tan (90^0-\theta = \cot\theta)$ and $\cot\theta\times \tan \theta =1$
$=1$

Hence proved.

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