Question

hey mate prove it please ​

Answer 1

SOLUTION:-

Given:

⚫x=cy +bz

⚫y=az+ cx

⚫z=bx + ay

To prove:

$\frac{ {x}^{2} }{1 - {a}^{2} } = \frac{ {y}^{2} }{1 - {b}^{2} } = \frac{ {z}^{2} }{1 - {c}^{2} }$

Proof:

Putting the value of x in y;

$y = az + c(cy + bz) \\ \\ y = z(a + bc) + {c}^{2y} \\ \\ y(1 - {c}^{2} ) = z(a + bc) \\ \\ (a + bc) = \frac{y(1 - {c}^{2} )}{z} .............(1)$

Similarly, putting equation (1) in z, we get;

$(a + bc) = \frac{z(1 - {b}^{2} )}{y}................(2)$

Equating equation (1) & (2), w get;

$\frac{ {y}^{2} }{(1 - {b}^{2} )} = \frac{ {z}^{2} }{(1 - {c}^{2} )} ...............(3)$

Similarly, by solving z & x by using value of y, we get;

$\frac{ {x}^{2} }{(1 - {a}^{2}) } = \frac{ {z}^{2} }{(1 - {c}^{2} )} ..............(4)$

Equation by (3) & (4), we get;

$= > \frac{ {x}^{2} }{(1 - {a}^{2} )} = \frac{ {y}^{2} }{(1 - {b}^{2}) } = \frac{ {z}^{2} }{(1 - {c}^{2} )}$

Hence,

Proved.

Hope it helps ☺️

Answer 2

Let us eliminate z from the twoequations as follows

X= cy + by

x= cy + b(bx+ay)

x= cy+b²x+aby

And now by multiplying each term ofthe x so we get ..

x ^{2} = cxy + b ^{2} x ^{2} + abxy.........(1)x2=cxy+b2x2+abxy.........(1)

And similarly

y = az + bxy=az+bx

y = a(bx + ay) + cxy=a(bx+ay)+cx

y = abx + a^{2} y + cxy=abx+a2y+cx

And by multiplying each term of y weget

y ^{2} = abxy + a ^{2} y^{2} + cxy..........(2)y2=abxy+a2y2+cxy..........(2)

Subtracting 2 from 1 we get

x²-y² = b²x²-a²y²

x²-b²x²=y²-a²y

x²(1-b²) = y²(1-a²)

$\frac{x ^{2} }{1 - a ^{2} } = \frac{b ^{2} }{1 - b ^{2} }1−a2x2=1−b2b2$

$\begin{lgathered}similarly \: we \: can \: eliminate \: y \: nd \: z \: from \: the \: equations \: to \: \: get \: \\ \\ \\\end{lgathered}similarlywecaneliminateyndzfromtheequationstoget</p><p>\frac{y ^{2} }{1 - b^{2} } = \frac{z ^{2} }{1 - c^{2} }1−b2y2=1−c2z2$

Hence proved

$\frac{x ^{2} }{1 - a ^{2} } = \frac{y ^{2} }{1 - b ^{2} } = \frac{z ^{2} }{1 - c ^{2} }1−a2x2=1−b2y2=1−c2z2$

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