This sample note is about Simple Problems in Mathematics.
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The Learning Hall Mathematics Class Oct 14, 2018 Out of 20 members in a family, 11 like to take tea and 14 like coffee. Assume that each one likes at least one of the two drinks. How many like 1. A. a. b. c. Let T = Both tea & coffee Only tea and not coffee Only coffee and not tea ? Set of members who like tea Set of members who like coffee By the problem, n (T) = 11 n (C) = 14 n (TUC) = 20 TUC c TnC a. or, Or, Or, We know, n (TUC) = n (T) + n (C) -n (Tnc) 20 11 + 14 —X = 25 -20=5 x n (Tnc) = 5 5 Members like both Tea & Coffee
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b. We know, n (TncC) = n (T) - n (Tnc) or, n (TnCC) = 11-5=6 c. 6 Members like Tea but not Coffee We know, n (cnTC) = n (C) - n (Tnc) 2. A. or, n 14-5=9 9 Members like Coffee but not Tea Find the values of the following a. Sin2300+Sin2450+ Sin2900 b. 3Sin2300+2tan2600-5 cos2450 a. b. Sin2300 + Sin2450+ Sin2600+ Sin2900 = 5/2 3Sin2300 + 2tan2600-5 cos2450 = + (1/12)2 = 3(1/4) + (1/2) = 3/4 + 6-5 / 2 = 17/4
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3. If tan x = b/a, then find the value of VI (a+b/a-b) + J (a-b/a+b) A. Now, Expression = VI (a+b/a-b) + (a-b/a+b) Dividing both numerators and denominator by Ja Expression = VI (l+b/a)/(l-b/a) + VI (I-b/a)/(l+b/a) = J ( 1+ tanx)/(l-tanx) + 1- tanx)/(l + tanx) = {(1+tanx) + (I-tanx)}/J (I-tan2x) = 2/1 (I-tan2x) = 2/1 (I-sin2x/cos2x) = 2 cosx/J(cos2x-sin2x) = 2 cosx/Jcos2x
Discussion
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