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S = { 1,2,3,4......,366}
n (S) = 366
A is event of leap year 53
A = { } n (A) = 0
P (A) = n (A) / n ( S)
= 0/ 366
so, only once 53 rd day will come
n (S) = 366
A is event of leap year 53
A = { } n (A) = 0
P (A) = n (A) / n ( S)
= 0/ 366
so, only once 53 rd day will come
Aarushiparadkar:
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Answered by
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Step-by-step explanation:
(1+ cotA — cosecA)(1+ tanA + secA)
1 + cotA — cosecA + tanA + cotA tanA — cosecA tanA + secA + cotA secA -cosecA secA
1 +cotA — cosecA + tanA +1- secA + secA+ cosecA — cosecA secA
1 +(cotA + tanA) +1 — cosecA secA
2 + cosecA secA — cosecA secA
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