If cot θ + cos θ = m and cot θ - cos θ = n, then prove that (m² - n²)² = 16 mn.

Given,

cot θ + cos θ = m ….(i)
cot θ - cos θ = n ….(ii)

Adding (i) and (ii) we get,

⇒ m + n = cot θ + cos θ + cot θ - cos θ
⇒ m + n = 2 cot θ
⇒ 2 cot θ = m + n
⇒ cot θ = $\dfrac{m + n}{2}$.

∴ tan θ = $\dfrac{2}{m + n}$ ….(iii)

Subtracting (ii) from (i) we get,

m - n = cot θ + cos θ - cot θ + cos θ
m - n = 2 cos θ
cos θ = $\dfrac{m - n}{2}$.

∴ sec θ = $\dfrac{2}{m - n}$ ….(iv)

Squaring and subtracting (iii) from (iv),

$\Rightarrow \text{sec}^2 \text{ θ} - \text{tan}^2 \text{ θ} = \Big(\dfrac{2}{m - n}\Big)^2 - \Big(\dfrac{2}{m + n}\Big)^2 \\[1em] \Rightarrow 1 = \dfrac{4}{(m - n)^2} - \dfrac{4}{(m + n)^2} \\[1em] \Rightarrow 4\Big[\dfrac{1}{(m - n)^2} - \dfrac{1}{(m + n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{(m + n)^2 - (m - n)^2}{(m + n)^2(m - n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{m^2 + n^2 + 2mn - m^2 - n^2 + 2mn}{(m + n)(m - n)(m + n)(m - n)} \Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{\cancel{m^2} + \cancel{n^2} + 2mn - \cancel{m^2} - \cancel{n^2} + 2mn}{(m^2 - n^2)(m^2 - n^2)} \Big] = 1 \\[1em] \Rightarrow 4 \times \dfrac{4mn}{(m^2 - n^2)^2} = 1 \\[1em] \Rightarrow 16 mn = (m^2 - n^2)^2.$

Hence, proved that (m² - n²)² = 16 mn.