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    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 < 90o
    A 4 Correct Answer Incorrect Answer
    B 2 Correct Answer Incorrect Answer
    C 1 Correct Answer Incorrect Answer
    D 3 Correct Answer Incorrect Answer

    Solution

    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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