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Question: Show that : \(\begin{gathered} \left( i \right)\tan {48^ \circ }\tan {23^ \circ }\tan {42^ \ci...

Show that :
(i)tan⁡48∘tan⁡23∘tan⁡42∘tan⁡67∘=1 (ii)cos⁡38∘cos⁡52∘−sin⁡38∘sin⁡52∘=0 \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}

Explanation

Solution

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

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