If - -1z 1z - x-y z2 - y-z x2 1x 1x -y y-z x2z x-2y-z xz - x-y xz2then

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Properties of Composite Function

if Δ =1z 1...

Question

if
$Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{z} & \frac{1}{z} & \frac{- (x + y)}{z^{2}} \frac{- (y + z)}{x^{2}} & \frac{1}{x} & \frac{1}{x} \frac{- y (y + z)}{x^{2} z} & \frac{(x + 2 y + z)}{x z} & \frac{- (x + y)}{x z^{2}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
then

Δ

is independent of

x

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Δ

is independent of

y

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Δ

is independent of

z

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Δ = 0

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∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{z} & \frac{1}{z} & - \frac{(x + y)}{z^{2}} - \frac{(y + z)}{x^{2}} & \frac{1}{x} & \frac{1}{x} - \frac{y (y + z)}{x^{2} z} & \frac{x + 2 y + z}{x z} & - \frac{y (x + y)}{x z^{2}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{z} & \frac{1}{z} & - \frac{(x + y)}{z^{2}} - \frac{(y + z)}{x^{2}} & \frac{1}{x} & \frac{1}{x} - \frac{y (y + z)}{x^{2} z} & \frac{x + 2 y + z}{x y} & - \frac{y (x + y)}{x z^{2}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣

(\neq 1)

Δ

⎡ ⎢ ⎣ \begin{matrix} 1 & l o g_{x} y & l o g_{x} z l o g_{y} x & 1 & l o g_{y} z s i n (x + y) & - c o s (x + y) & s i n^{2} z \end{matrix} ⎤ ⎥ ⎦

Δ

x = sin (α - β) sin (γ - δ)

y = sin (β - γ) sin (α - δ)

z = sin (γ - α) sin (β - δ)

x + y + z = 0

2 sin A sin B = cos (A - B) + cos (A + B)

\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}

\frac{x - b + c}{δ - λ} = \frac{y - b}{β} = \frac{z - b - c}{β + γ}

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