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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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