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11 January, 09:52

1 + tanx / 1 + cotx = 2

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  1. 11 January, 10:31
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    x = tan^ (-1) ((i sqrt (3)) / 2 + 1/2) + π n_1 for n_1 element Z

    or x = tan^ (-1) ( - (i sqrt (3)) / 2 + 1/2) + π n_2 for n_2 element Z

    Step-by-step explanation:

    Solve for x:

    1 + cot (x) + tan (x) = 2

    Multiply both sides of 1 + cot (x) + tan (x) = 2 by tan (x):

    1 + tan (x) + tan^2 (x) = 2 tan (x)

    Subtract 2 tan (x) from both sides:

    1 - tan (x) + tan^2 (x) = 0

    Subtract 1 from both sides:

    tan^2 (x) - tan (x) = - 1

    Add 1/4 to both sides:

    1/4 - tan (x) + tan^2 (x) = - 3/4

    Write the left hand side as a square:

    (tan (x) - 1/2) ^2 = - 3/4

    Take the square root of both sides:

    tan (x) - 1/2 = (i sqrt (3)) / 2 or tan (x) - 1/2 = - (i sqrt (3)) / 2

    Add 1/2 to both sides:

    tan (x) = 1/2 + (i sqrt (3)) / 2 or tan (x) - 1/2 = - (i sqrt (3)) / 2

    Take the inverse tangent of both sides:

    x = tan^ (-1) ((i sqrt (3)) / 2 + 1/2) + π n_1 for n_1 element Z

    or tan (x) - 1/2 = - (i sqrt (3)) / 2

    Add 1/2 to both sides:

    x = tan^ (-1) ((i sqrt (3)) / 2 + 1/2) + π n_1 for n_1 element Z

    or tan (x) = 1/2 - (i sqrt (3)) / 2

    Take the inverse tangent of both sides:

    Answer: x = tan^ (-1) ((i sqrt (3)) / 2 + 1/2) + π n_1 for n_1 element Z

    or x = tan^ (-1) ( - (i sqrt (3)) / 2 + 1/2) + π n_2 for n_2 element Z
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