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

Step-by-step explanation:

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