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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=2cos⁡θ 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    
​cos⁡2θ=2cos⁡2θ−1\cos 2 \theta = 2\cos^2 \theta -1​​
Solution:  
​x+1x=2cos⁡θ x+\frac{1}{x}=2 cos⁡θ​,  
squaring both side;
we get :
​(x+1x)2=(2cos⁡θ)2 x2+1x2+2=4cos⁡2θ x2+1x2=4cos⁡2θ−2 x2+1x2=2(2cos⁡2θ−1) x2+1x2=2cos⁡2θ\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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