if sin(A+B)=1 , sin(A-B)=1/2 then tan(A+2B) tan(2A+B) is equal to?
A.1
B.-1
C.0
D.None of these
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a 1 is the right ans
Put S = sin, C = cos, T = tan. Then S(a+b) = 1---> a+b = 90...[1], S(a-b) = (1/2)---> a-b = 30...[2], {[1]+[2]}--->
a = 60, then b = 30. Then a+2b = 120 and T(120) = -T(60), 2a+b = 150 and T(150) = -T(30). Now T(30)T(60) =
(1/rt3)(rt3) = 1 so tan(a+2b)tan(2a+b) = 1.
Put S = sin, C = cos, T = tan. Then S(a+b) = 1---> a+b = 90...[1], S(a-b) = (1/2)---> a-b = 30...[2], {[1]+[2]}--->
a = 60, then b = 30. Then a+2b = 120 and T(120) = -T(60), 2a+b = 150 and T(150) = -T(30). Now T(30)T(60) =
(1/rt3)(rt3) = 1 so tan(a+2b)tan(2a+b) = 1.
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