If x(1+ )=4 m and x(1- )=4 n,prove that (m^2-n^2)^2=mn.

Question

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

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Answer

Let, $\cot\text{x}(1+\sin\text{x})=4\text{m}\cdots\text{(i)}$ and, $\cot\text{x}(1-\sin\text{x})=4\text{n}\cdots\text{(ii)}$ To show: $\text{(m}^2-\text{n}^2)^2=\text{mn}$ From (i) and (ii), we get $\text{m}=\frac{\cot\text{x}(1+\sin\text{x})}{4}\&\text{ n}=\frac{\cot\text{x}(1-\sin\text{x})}{4}$ $\text{L.H.S}=(\text{m}^2-\text{n}^2)^2$ $=((\text{m}+\text{n})(\text{m}-\text{n}))^2$ $=(\text{m}+\text{n})^2(\text{m}-\text{n})^2$ $=\Big(\frac{\cot\text{x}(1+\sin\text{x})+\cot\text{x}(1-\sin\text{x})}{4}\Big)^2\times\Big(\frac{\cot\text{x}(1+\sin\text{x})-\cot\text{x}(1-\sin\text{x})}{4}\Big)^2$ $=\Big(\frac{\cot\text{x}(1+\sin\text{x}+1-\sin\text{x})}{4}\Big)\times\Big(\frac{\cot\text{x}(1+\sin\text{x}-1-\sin\text{x})}{4}\Big)^2$ $=\frac{\cot^2\text{x}}{16}\times4\times\frac{\cot^2\text{x}}{16}\times\sin^2\text{x}$ $\frac{\cot^2\text{x}}{16}\times\frac{\cos^2\text{x}}{\sin^2\text{x}}\sin^2\text{x}$ $\Big[\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\Big]$ $=\frac{\cot\text{x}}{4}\times\frac{\cot\text{x}}{4}\times(1-\sin^2\text{x})$ $\Big[\because\cos^2\text{x}=1=\sin^2\text{x}\Big]$ $=\frac{\cot\text{x}(1+\sin\text{x})}{4}\times\frac{\cot\text{x}(1-\sin\text{x})}{4}$ $=\text{mn}.$

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