Show that:
(i) $tan\ 48^o\ tan\ 23^o\ tan\ 42^o\ tan\ 67^o = 1$
(ii) $cos\ 38^o\ cos\ 52^o - sin\ 38^o\ sin\ 52^o = 0$

To do:

We have to show that 

(i) $tan\ 48^o\ tan\ 23^o\ tan\ 42^o\ tan\ 67^o = 1$.

(ii) $cos\ 38^o\ cos\ 52^o - sin\ 38^o\ sin\ 52^o = 0$.

Solution:  

(i) We know that,

$tan\ (90^{\circ}- \theta) = cot\ \theta$

$tan\ \theta \times \cot\ \theta=1$

Therefore,

LHS $=\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=\tan (90^{\circ}-42^{\circ})\tan 23^{\circ}\tan 42^{\circ}\tan (90^{\circ}-23^{\circ})$

$=\tan 42^{\circ}\tan 23^{\circ}\cot 42^{\circ}\cot 23^{\circ}$

$=(\tan 42^{\circ}\cot 42^{\circ})(\tan 23^{\circ}\cot 23^{\circ})$

$=1\times1$

$=1$

$=$ RHS

Hence proved.

(ii) We know that,

$\cos\ (90^{\circ}- \theta) = \sin\ \theta$

$\sin\ (90^{\circ}- \theta) = \cos\ \theta$

Therefore,

LHS $=cos\ 38^o\ cos 52^o - sin\ 38^o sin\ 52^o$

$= cos\ 38^o cos\ (90^o - 38^o) - sin\ 38^o sin\ (90^o - 38^o)$

$= cos\ 38^o sin\ 38^o - sin\ 38^o cos\ 38^o$

$=0$

$=$ RHS

Hence proved. 

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Updated on: 10-Oct-2022

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