Question
If sec(2A + B) = 2 and cos(A + B) = (1/√2) , find the
value of {sec 2B + tan (A + B) } given that 0o < A, B < 90oSolution
We have, sec(2A + B) = 2
Or, sec(2A + B) = sec 60o
Or, 2A + B = 60o -------- (I)
And, cos(A + B) = (1/√2)
Or, cos(A + B) = cos 45o
Or, A + B = 45o ------- (II)
On subtracting equation (II) from (I) ,
We have, (2A + B) - (A + B) = 60o - 45o
Or, 'A' = 15o
Put the value of 'A' = 15o in equation (1) ,
So, 2 X 15o + B = 60o
Or, 30o + B = 60o
Or, 'B' = 30o
Therefore, required value = sec (2 X 30o) + tan (15o+30o)
= sec 60o + tan 45o = 2 + 1 = 3
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