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

Answer:

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} }

1−a

2

x

2

=

1−b

2

y

2

=

1−c

2

z

2

Proof:

Putting the value of x in y;

\begin{gathered}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)\end{gathered}

y=az+c(cy+bz)

y=z(a+bc)+c

2y

y(1−c

2

)=z(a+bc)

(a+bc)=

z

y(1−c

2

)

.............(1)

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

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

y

z(1−b

2

)

................(2)

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

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

(1−b

2

)

y

2

=

(1−c

2

)

z

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)

(1−a

2

)

x

2

=

(1−c

2

)

z

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} )}=>

(1−a

2

)

x

2

=

(1−b

2

)

y

2

=

(1−c

2

)

z

2

Hence,

Proved.