Question

If $\cot x(1+\sin x)=4m$ and $\cot x(1-\sin x)=4n$, prove that $(m^2-n^2{)}^2=mn$.

Answer 1

$\to m^2-n^2=\frac{1}{16}{\cot }^{2}x[{(1+sinx)}^2-{(1-sinx)}^2]$

$\to m^2-n^2=\frac{{\cot }^{2}x}{16}\times 4sinx=\frac{sinx}{4}{\cot }^{2}x$

$\to {(m^2-n^2)}^2=\frac{{\sin }^{2}x{\cot }^{4}x}{16}=LHS$

$\to \frac{{\sin }^{2}x}{16}\frac{{\cos }^{2}x}{{\sin }^{2}x}{\cot }^{2}x=\frac{{\cot }^{2}x}{16}{\cos }^{2}x=LHS$

$\to mn=\frac{1}{16}{\cot }^{2}x(1-{\sin }^{2}x)=\frac{1}{16}{\cot }^{2}x{\cos }^{2}x=RHS$

LHS=RHS