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    If x+1/x=2 cos⁡θ,  then what is the value of x2+1x2x^2+\frac{1}{x^2}x2+x21​​ ?
    Question

    If x+1/x=2 cos⁡θ,  then what is the value of x2+1x2x^2+\frac{1}{x^2}​ ?

    A.

    sin⁡2θ

    B.

    2 cos⁡2θ

    C.

    2 sin⁡2θ

    D.

    cos⁡2θ

    Correct option is B

    Given: 
    x+1x=2cosθ x+\frac{1}{x}=2 cos⁡θ​, 
    To, find value of x2+1x2x^2+\frac{1}{x^2} 
    Formula Used: 
    (a+b)2=a2+b2+2ab(a +b)^2 = a^2 + b^2 + 2ab    
    cos2θ=2cos2θ1\cos 2 \theta = 2\cos^2 \theta -1​​
    Solution:  
    x+1x=2cosθ x+\frac{1}{x}=2 cos⁡θ​,  
    squaring both side;
    we get :
    (x+1x)2=(2cosθ)2 x2+1x2+2=4cos2θ x2+1x2=4cos2θ2 x2+1x2=2(2cos2θ1) x2+1x2=2cos2θ\left( x+\frac{1}{x} \right)^2 = (2 \cos \theta)^2 \\ \ \\x^2 + \frac{1}{x^2 }+2 = 4 \cos^2 \theta \\ \ \\x^2 + \frac{1}{x^2 } = 4 \cos^2 \theta -2 \\ \ \\x^2 + \frac{1}{x^2 } = 2(2\cos^2 \theta - 1) \\ \ \\x^2 + \frac{1}{x^2 } = 2\cos 2 \theta ​​​

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