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  • Show that tan 48^o tan 23^o tan 42^o tan 67^o=1

2.    Show that :

 (i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1

Answers (1)

best_answer

\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.

Posted by

manish

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