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    If cot⁡A + cos⁡A = m and cot⁡A - cos⁡A = n, then:
    Question

    If cot⁡A + cos⁡A = m and cot⁡A - cos⁡A = n, then:

    A.

    m2+n2=4mnm^2+n^2=4mn​​

    B.

    (m2n2)2=16mn(m^2-n^2 )^2=16mn​​

    C.

    (m2n2)2=4mn(m^2-n^2 )^2=4mn​​

    D.

    m2n2=16mnm^2-n^2=16mn​​

    Correct option is B

    ​Given:

    m = cot⁡A + cos⁡A

    n = cot⁡A − cos⁡A 

    Formula Used:

    m2n2=(m+n)(mn)m^2 - n^2 = (m + n)(m - n)

    Solution:

    Add m and n: 

    m + n = cot⁡A + cos⁡A + (cot⁡A − cos⁡A) 

    m + n = 2cot⁡A 

    Subtract m and n: 

    m - n = cot⁡A + cos⁡A - (cot⁡A − cos⁡A) 

    m − n = 2cos⁡A

    Square difference:

    (m2n2)2 =(2cotA2cosA)2 =16cotAcosA =16 mn(m^2 - n^2)^2 \\ \ \\= (2 \cot A \cdot 2 \cos A)^2\\ \ \\ = 16 \cot A \cos A \\ \ \\= 16 \ mn

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