Question

If $3x=cosecA$ and $3/x = cotA$, then $3(x^{2} -1/x^{2} )$

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ $\displaystyle\sf 3x = cosec \ A$

✭ $\displaystyle\sf \dfrac{3}{x} = cot \ A$

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ $\displaystyle\sf 3(x^2-{\dfrac{1}{x}}^2)$

$\displaystyle\large\underline{\sf\gray{Solution}}$

Here we shall first find the values of x & ¹/x and then simply substitute in the given equation [3(x²-{¹/x}²]

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

⨳ $\displaystyle\sf 3x = cosec \ A$

⨳ $\displaystyle\sf \purple{x = \dfrac{cosec \ A}{3}}\:\:\: -eq(1)$

Similarly,

»» $\displaystyle\sf \dfrac{3}{x} = cot \ A$

»» $\displaystyle\sf \dfrac{1}{x} = cot \ A \times 3$

»» $\displaystyle\sf \green{\dfrac{1}{x} = \dfrac{cot \ A}{3}}\:\:\: -eq(2)$

So then,

➳ $\displaystyle\sf 3\bigg\lgroup x^2-{\dfrac{1}{x}}^2\bigg\rgroup$

➳ $\displaystyle\sf 3\bigg\lgroup (\dfrac{cosec \ A}{3})^2-(\dfrac{cot \ A}{3})^2\bigg\rgroup$

[From eq(1) & eq(2)]

➳ $\displaystyle\sf 3\bigg\lgroup (\dfrac{cosec^2 \ A}{9})-(\dfrac{cot^2 \ A}{9})\bigg\rgroup$

➳ $\displaystyle\sf 3\bigg\lgroup \dfrac{cosec^2 \ A - cot^2 \ A}{9}\bigg\rgroup$

$\displaystyle\small\sf \qquad\quad \because cosec^2 \ A-cot^2 \ A = 1$

➳ $\displaystyle\sf 3\times \dfrac{1}{9}$

➳ $\displaystyle\sf \pink{\dfrac{1}{3}}$

$\displaystyle\sf \therefore\:\underline{\sf The \ value \ of \ 3(x^2-{\dfrac{1}{x}}^2) \ is \ \dfrac{1}{3}}$

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Answer 2

$\Large{\underline{\underline{\green{\sf Given:}}}}$

$\sf\star 3x = \cosec A$

$\sf\star \dfrac{3}{x} = \cot A$

$\Large{\underline{\underline{\green{\sf Find:}}}}$

$\sf\star 3( {x}^{2} - \dfrac{1}{ {x}^{2} } )$

$\Large{\underline{\underline{\green{\sf Solution:}}}}$

Here, it is given

$\sf\to 3x = \cosec A$

$\sf\to x = \dfrac{\cosec A}{3}.......(1)$

$\sf\mapsto \dfrac{3}{x} = \cot A$

$\sf\mapsto \dfrac{1}{x} = \dfrac{\cot A}{3}........(2)$

we, have

$\sf\hookrightarrow 3( {x}^{2} - \dfrac{1}{ {x}^{2} } )$

Use eq(1) and (2)

$\sf\hookrightarrow 3( \bigg({\dfrac{\cosec A}{3} \bigg)}^{2} - \dfrac{1}{ {x}^{2} } )$

$\sf\hookrightarrow 3( \bigg({\dfrac{\cosec A}{3} \bigg)}^{2} - {\bigg(\dfrac{\cot A}{3}\bigg)}^{2} )$

$\sf\hookrightarrow 3( \bigg(\dfrac{{\cosec}^{2} A}{3} \bigg) - \bigg(\dfrac{{\cot}^{2} A}{3}\bigg) )$

$\sf\hookrightarrow 3 \bigg(\dfrac{{\cosec}^{2} A - {\cot}^{2} A}{9} \bigg)$

$\sf\hookrightarrow 3 \bigg(\dfrac{1}{9} \bigg) \qquad \bigg[ {\cosec}^{2} A - {\cot}^{2}A = 1 \bigg]$

$\sf\hookrightarrow \cancel{\dfrac{3}{9}} = \dfrac{1}{3}$

$\sf\hookrightarrow \dfrac{1}{3}$

$\small{\sf \therefore 3( {x}^{2} - \dfrac{1}{ {x}^{2} } ) = \dfrac{1}{3} }$