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Question : 21 of 160
Marks:
+1,
-0
Solution:
‌cos‌13∘⋅sin‌17∘⋅sin‌21∘⋅cos‌47∘‌=(cos‌13∘‌cos‌47∘)⋅(sin‌17∘sin‌21∘)‌=‌[cos(34∘)+cos(60∘)]⋅‌[cos(4∘)−cos(38∘)]‌[∵sin‌Asin‌B=‌[cos(A−B)−cos(A+B)‌ and ‌cos‌A‌cos‌B=‌[cos(A−B)+cos(A+B)]..‌=‌[cos‌34∘+‌][cos‌4∘−cos‌38∘]‌=‌[cos‌34∘‌cos‌4∘−cos‌34∘‌cos‌38∘+‌‌cos‌4∘−‌‌cos‌38∘]‌=‌[‌{cos(30∘)+cos(38∘)}−‌{cos(4∘)+cos(72∘)}+‌‌cos‌4∘−‌‌cos‌38∘]‌=‌[‌(‌+cos‌38∘)−‌(cos‌4∘+cos‌72∘)+‌‌cos‌4∘−‌‌cos‌38∘]‌=‌[‌−cos‌72∘]=‌−‌‌=‌−‌×‌(∵cos‌72∘=‌)‌=‌−‌=‌=‌(2√3−√5+1)
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