Question

If Sinϴ = 2x and xcosϴ = 2, 0⁰ < ϴ < 90⁰ , find the value of 2 ( x^2 + 1 / x^2 ).

Answer 1

Answer:

2(x² + 1/x ) = 1/2

Step-by-step explanation:

Given,

sinϴ = 2x and xcosϴ = 2 , 0<ϴ<90

we hav to find,

2(x² + 1/x²)

as,

=> sinϴ = 2x , cosϴ = 2/x

on adding and squaring on both sides

=> (sinϴ + cosϴ)² = (2x + 2/x)²

=> sin²ϴ + cos²ϴ + 2sinϴcosϴ = 4x² + 4/x² + 2*2x*2/x

=> (sin²ϴ + cos²ϴ) + 2sinϴcosϴ = 4x² + 4/x² + 8

[ sin²ϴ + cos²ϴ = 1]

=> 1 + 2sinϴcosϴ = 4x² + 4/x² + 8

=> 1 - 8 + 2sinϴcosϴ = 4(x² + 1/x²)

=> 4(x² + 1/x²) = -7 + 2sinϴcosϴ

=> 2(x² + 1/x²) = (sinϴcosϴ - 7/2)

but sinϴ = 2x & cosϴ = 2/x

sinϴcosϴ = 2x × 2/x

sinϴcosϴ = 4

2(x² + 1/x²) = 4 - 7/2

2(x² + 1/x²) = (8-7)/2

2(x² + 1/x²) = 1/2

thanks

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Answer 2

$\boxed{2(x^{2}+\dfrac{1}{x^{2}})=\dfrac{1}{2}}$

Given data:

$sin\theta=2x$ and $x\:cos\theta=2$, $0^{\circ}<\theta<90^{\circ}$

To find:

The value of $2(x^{2}+\dfrac{1}{x^{2}})$

Concept to be used:

Before we solve this problem, we must know

$sin^{2}\theta+cos^{2}\theta=1$

Step-by-step explanation:

Here, $sin\theta=2x$ and $x\:cos\theta=2$

$\Rightarrow sin\theta=2x$ and $cos\theta=\dfrac{2}{x}$

Also, $sin^{2}\theta+cos^{2}\theta=1$

$\Rightarrow (2x)^{2}+(\dfrac{2}{x})^{2}=1$

$\Rightarrow 4x^{2}+\dfrac{4}{x^{2}}=1$

$\Rightarrow 4(x^{2}+\dfrac{1}{x^{2}})=1$

$\Rightarrow 2(x^{2}+\dfrac{1}{x^{2}})=\dfrac{1}{2}$

Thus the value of $2(x^{2}+\dfrac{1}{x^{2}})$ is $\dfrac{1}{2}$.

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