Question

Show that :
$\begin{gathered}
  \left( i \right)\tan {48^ \circ }\tan {23^ \circ }\tan {42^ \circ }\tan {67^ \circ } = 1 \\
  \left( {ii} \right)\cos {38^ \circ }\cos {52^ \circ } - \sin {38^ \circ }\sin {52^ \circ } = 0 \\
\end{gathered} $

Answer 1

Hint:In this question use some basic trigonometric conversions like $\tan \left( {90 - \theta } \right) = \cot \theta $,$\sin \left( {90 - \theta } \right) = \cos \theta $ , $\cos \left( {90 - \theta } \right) = \sin \theta $,$\cot \left( {90 - \theta } \right) = \tan \theta $, $\cot \theta = \dfrac{1}{{\tan \theta }}$.

Complete step-by-step answer:
According to the question
(i) We have $\tan {48^ \circ }\tan {23^ \circ }\tan {42^ \circ }\tan {67^ \circ } = 1$
$LHS = $$\tan {48^ \circ }\tan {23^ \circ }\tan {42^ \circ }\tan {67^ \circ }$
           $
   = \tan \left( {{{90}^ \circ } - {{42}^ \circ }} \right)\tan {23^ \circ }\tan {42^ \circ }\tan \left( {{{90}^ \circ } - {{23}^ \circ }} \right) \\
   = \cot {42^ \circ }\tan {42^ \circ }\tan {23^ \circ }\cot {23^ \circ } \\
   = \cot {42^ \circ } \times \dfrac{1}{{\cot {{42}^ \circ }}} \times \tan {23^ \circ } \times \dfrac{1}{{\tan {{23}^ \circ }}} = 1 \\
 $
            $ = R.H.S.$ Hence , Proved
(ii) We have $\cos {38^ \circ }\cos {52^ \circ } - \sin {38^ \circ }\sin {52^ \circ } = 0$
$LHS = \cos {38^ \circ }\cos {52^ \circ } - \sin {38^ \circ }\sin {52^ \circ }$
           $
   = \cos \left( {{{90}^ \circ } - {{52}^ \circ }} \right)\cos {52^ \circ } - \sin \left( {{{90}^ \circ } - {{52}^ \circ }} \right)\sin {52^ \circ } \\
   = \sin {52^ \circ }\cos {52^ \circ } - \cos {52^ \circ }\sin {52^ \circ } \\
   = 0 = RHS \\
$
Hence proved .

Note: It is always advisable to remember some basic conversions while involving trigonometric questions.Students should remember the trigonometric identities and formulas for solving these types of questions.