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If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° B, find A and B. Q 3.

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What is the best solution of ncert class 10 introduction of trigonometry question . Please find out the best and easy solution of the exercise 8.2 question number 3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

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  1. tan (A + B) = √3

    Since √3 = tan 60°

    Now substitute the degree value

    ⇒ tan (A + B) = tan 60°

    (A + B) = 60° … (i)

    The above equation is assumed as equation (i)

    tan (A – B) = 1/√3

    Since 1/√3 = tan 30°

    Now substitute the degree value

    ⇒ tan (A – B) = tan 30°

    (A – B) = 30° … equation (ii)

    Now add the equation (i) and (ii), we get

    A + B + A – B = 60° + 30°

    Cancel the terms B

    2A = 90°

    A= 45°

    Now, substitute the value of A in equation (i) to find the value of B

    45° + B = 60°

    B = 60° – 45°

    B = 15°

    Therefore A = 45° and B = 15°

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