The equation sin 2-x2-y2-2 x y- x- y-0 is possible if

The correct option is A x = y Given, sin^2 θ = x^2 + y^2/2xy ∴ x, y must have same sign Also, 0≤ x^2+y^2 ≤ 2xy ⇒ 0 ≤ x^2+y^2 2xy ≤ 0 ⇒ 0 ≤ (x y)^2 ≤ 0 ⇒ x=y ...

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Standard XII

Mathematics

Range

The equation ...

Question

The equation $s i n^{2} θ = \frac{x^{2} + y^{2}}{2 x y}, x, y \neq 0$ is possible if

$x = y$

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$x = - y$

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$2 x = y$

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$4 x = y$

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Solution

The correct option is A
$x = y$

Given, ${sin}^{2} θ = \frac{x^{2} + y^{2}}{2 x y}$

$∴ x, y$ must have same sign

Also, $0 \leq x^{2} + y^{2} \leq 2 x y$

$\Rightarrow 0 \leq x^{2} + y^{2} - 2 x y \leq 0$

$\Rightarrow 0 \leq (x - y)^{2} \leq 0$

$\Rightarrow x = y$

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The equation sin2θ=x2+y22xy, x, y≠0 is possible if

The correct option is A x=y Given, sin2θ=x2+y22xy ∴x, y must have same sign Also, 0≤x2+y2≤2xy ⇒0≤x2+y2−2xy≤0 ⇒0≤(x−y)2≤0 ⇒x=y

The equation $s i n^{2} θ = \frac{x^{2} + y^{2}}{2 x y}, x, y \neq 0$ is possible if

The correct option is A
$x = y$

Given, ${sin}^{2} θ = \frac{x^{2} + y^{2}}{2 x y}$

$∴ x, y$ must have same sign

Also, $0 \leq x^{2} + y^{2} \leq 2 x y$

$\Rightarrow 0 \leq x^{2} + y^{2} - 2 x y \leq 0$

$\Rightarrow 0 \leq (x - y)^{2} \leq 0$

$\Rightarrow x = y$