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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions3 Marks
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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