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Q2. Show that
(i) \( tan 48° tan 23° tan 42° tan 67° \ = \ 1 \)

(ii) \( cos 38° cos 52° \ - \ sin 38° sin 52 ° \ = \ 0 \)

Answer :


(i) \( tan 48° tan 23° tan 42 ° tan 67° \ = \ tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°\ = \ cot 42° cot 67° tan 42° tan 67° \)    [∵ \( tan(90° - \theta) \ = \ cot\theta \) ]

\( = \ \frac{1}{tan42°} × \frac{1}{tan67°} × tan42°tan67° \ = \ 1 \)

(ii) \( cos 38°cos 52°– sin 38°sin 52° \ = \ cos(90° – 52°)cos 52° – sin(90°– 52°)sin 52° \)

\( = \ sin 52° cos 52° - cos 52° sin 52° \ = \ 0 \)    [∵ \( cos(90°-\theta) \ = \ sin\theta \) and \( sin(90°-\theta) \ = \ cos\theta \) ]