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Question : 20 of 160
Marks:
+1,
-0
Solution:
‌‌ We have, ‌‌| 1 |
| sin‌r∘⋅sin‌(r+1)∘ |
‌⇒‌‌‌| sin‌((r∘+1∘)−r∘) |
| sin‌r∘⋅sin‌(r∘+1∘) |
‌⇒‌‌‌| sin‌(r+1)∘‌cos‌r∘ |
| sin‌r∘⋅sin‌(r+1)∘ |
‌⇒‌‌∑r=1∘89cot‌r∘−cot(r+1)∘‌⇒‌[cot‌1∘−cot‌90∘]=‌
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