If cot x


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


Given :

$4 m=\cot x(1+\sin x)$ and $4 n=\cot x(1-\sin x)$

Multiplying both the equations:

$\Rightarrow 16 m n=\cot ^{2} x\left(1-\sin ^{2} x\right)$

$\Rightarrow 16 m n=\cot ^{2} x \cdot \cos ^{2} x$

$\Rightarrow m n=\frac{\cos ^{4} x}{16 \sin ^{2} x}$    ....(1)

Squaring the given equation :

$16 m^{2}=\cot ^{2} x(1+\sin x)^{2}$ and $16 n^{2}=\cot ^{2} x(1-\sin x)^{2}$

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

$\Rightarrow m^{2}-n^{2}=\frac{\cot ^{2} x \cdot \sin x}{4}$

Squaring both sides,

$\left(m^{2}-n^{2}\right)^{2}=\frac{\cot ^{4} x \cdot \sin ^{2} x}{16}$

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

From $(1)$ and $(2):$

$\left(m^{2}-n^{2}\right)^{2}=m n$

Hence proved.

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