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if tanx+cotx=2 then the value of tan^5x + cot^10x is

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  1. If tanx + cotx = 2
    sinx/cosx + cosx/sinx = 2
    (sin^2 x + cos^2 x)/(sinxcosx) = 2
    1/(sinxcosx) = 2
    2sinxcosx= 1
    sin 2x = 1
    2x = 90° for 0 ≤ x ≤ 180°
    x = 45° or π/4 radians
    we know tan 45° = 1 and cot 45° = 1

    then tan^5 x + cot^10 x
    = 1^5 + 1^10
    = 2


    There are other solutions for x
    they are x + k(180°) , where k is an integer
    but all such angles will fall in either quadrant I or III and both the tangent and cotangent would be +1
    so tan^5 x + cot^2 x = 2

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