If \[ \cot x(1+\sin x)=4m \] and \[ \cot x(1-\sin x)=4n \] Prove that \[ (m^2-n^2)^2=mn \]

Solution:

\[ 4m=\cot x(1+\sin x) = \frac{\cos x}{\sin x}(1+\sin x) \]

\[ 4n=\cot x(1-\sin x) = \frac{\cos x}{\sin x}(1-\sin x) \]

Multiplying,

\[ 16mn = \frac{\cos^2 x}{\sin^2 x}(1-\sin^2 x) \]

\[ = \frac{\cos^4 x}{\sin^2 x} \]

Now,

\[ 4(m+n) = \frac{\cos x}{\sin x}(2) \]

\[ m+n=\frac{\cos x}{2\sin x} \]

Also,

\[ 4(m-n) = \frac{\cos x}{\sin x}(2\sin x) \]

\[ m-n=\frac{\cos x}{2} \]

Therefore,

\[ m^2-n^2 = (m+n)(m-n) \]

\[ = \frac{\cos x}{2\sin x}\cdot\frac{\cos x}{2} \]

\[ = \frac{\cos^2 x}{4\sin x} \]

Squaring both sides,

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

\[ =mn \]

Hence proved.