If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]

#### Solution

Given:

\[4m = cotx\left( 1 + \sin x \right) and 4n = cot x\left( 1 - \sin x \right)\]

Multiplying both the equations:

\[ \Rightarrow 16mn = co t^2 x\left( 1 - \sin^2 x \right)\]

\[ \Rightarrow 16mn = co t^2 x . \cos^2 x\]

\[ \Rightarrow mn = \frac{\cos^4 x}{16 \sin^2 x} \left( 1 \right)\]

Squaring the given equation:

\[16 m^2 = co t^2 x \left( 1 + \sin x \right)^2 \text{ and }16 n^2 = co t^2 x \left( 1 - \sin x \right)^2 \]

\[ \Rightarrow 16 m^2 - 16 n^2 = co t^2 x\left( 4\sin x \right)\]

\[ \Rightarrow m^2 - n^2 = \frac{co t^2 x . \sin x}{4}\]

Squaring both sides,

\[ \left( m^2 - n^2 \right)^2 = \frac{co t^4 x . \sin^2 x}{16}\]

\[ \Rightarrow \left( m^2 - n^2 \right)^2 = \frac{\cos^4 x}{16 \sin^2 x} (2)\]

From (1) and (2):

\[ \left( m^2 - n^2 \right)^2 = mn\]

Hence proved.