Question

If 3x=cosecθ and 3/x=cotθ, then find the value of 3(x²-1/x²) PLEASE HELP BEST ANSWER WILL BE MARKED BRAINLIEST

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric identity that

$\sf{{ \cosec}^{2} \theta -{ \cot}^{2} \theta = 1 }$

GIVEN

$\displaystyle \: \sf{3x = \cosec \theta \: \: \: and \: \: \frac{3}{x} = \cot \theta \: \: \: }$

TO FIND

$\displaystyle \: \sf{3 \bigg( {x}^{2} - \frac{1}{ {x}^{2} } \bigg )\: }$

CALCULATION

$\displaystyle \: \sf{3x = \cosec \theta \: \: }$

$\implies\displaystyle \: \sf{x = \frac{1}{3} \cosec \theta } \: \: \: .....(1)$

Again

$\displaystyle \: \sf{ \frac{3}{x} = \cot \theta \: \: \: }$

$\implies \: \displaystyle \: \sf{ \frac{1}{x} = \frac{1}{3} \cot \theta \: \: \: } \: \: \: ......(2)$

Now

$\displaystyle \: \sf{3 \bigg[ {x}^{2} - \frac{1}{ {x}^{2} } \bigg] \: }$

$= \displaystyle \: \sf{3 \bigg[ { \bigg( \frac{1}{3} \cosec \theta \bigg)}^{2} - { \bigg( \frac{1}{3} \cot \theta \bigg)}^{2} \: \bigg] \: }$

$= \displaystyle \: 3 \times \frac{1}{9} \bigg( \sf{{ \cosec}^{2} \theta -{ \cot}^{2} \theta \bigg) }$

$= \displaystyle \: \frac{1}{3}$

RESULT

$\boxed{ \displaystyle \: \sf{ \: 3 \bigg[ {x}^{2} - \frac{1}{ {x}^{2} } \bigg] \: } = \frac{1}{3} \: \: \: }$

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Answer 2

$\underline{\textsf{Given:}}$

$\mathsf{3x=\,cosec\theta}$

$\mathsf{\dfrac{3}{x}=cot\,\theta}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}\;\mathsf{3(x^2-\dfrac{1}{x^2})}$

$\underline{\textsf{Solution:}}$

$\textsf{We know that, the identity}$

$\mathsf{cosec^2\theta-cot^2\theta=1}$

$\implies\mathsf{(3x)^2-(\dfrac{3}{x})^2=1}$

$\implies\mathsf{9x^2-\dfrac{9}{x^2}=1}$

$\implies\mathsf{9(x^2-\dfrac{1}{x^2})=1}$

$\implies\mathsf{3{\times}3(x^2-\dfrac{1}{x^2})=1}$

$\implies\boxed{\mathsf{3(x^2-\dfrac{1}{x^2})=\dfrac{1}{3}}}$

$\underline{\textsf{Answer:}}$

$\textsf{The value of}\;\mathsf{3(x^2-\dfrac{1}{x^2})}$

$\textsf{is}\;\mathsf{\dfrac{1}{3}}$

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