New Enroll in RBI Grade B New Batch - Starts on May 21

    Question

    If SecA + TanA = 2√2 + 9, then find the value of

    Sin A + Cos A.
    A (1+2√5)/9 Correct Answer Incorrect Answer
    B (2+2√5)/9 Correct Answer Incorrect Answer
    C (1+4√5)/9 Correct Answer Incorrect Answer
    D (2(1+2√5))/9 Correct Answer Incorrect Answer

    Solution

    Sec²A – Tan²A = 1 (SecA – TanA) (SecA + TanA) = 1 (SecA – TanA) (2√2 + 9) = 1 (SecA - TanA) = 1/(2√2 + 9) SecA – TanA = 9 - 2√2 ………………(i) SecA + TanA = 9 + 2√2 ……………… (ii) By solving the two equations we get, 2SecA = 18 SecA = 9 CosA = 1/9 Sin²A + Cos²A = 1 Sin²A + 1/81 = 1 Sin²A = 1 - 1/81 Sin²A = 80/81 SinA= (4√5)/9 SinA + CosA = (4√5)/9+ 1/9 SinA + CosA = (1+4√5)/9

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