I'm done with this question..Can anyone help me?

tan^-1(yz/xr) + tan^-1(zx/yr) +tan^-1(xy/zr) = pi/4 Prove that x^2+ y^2+ z^2=r^2

Hello, to do this sum there is an easiest formula, tan^-1x + tan^-1y + tan^-1 z = x+ y + z - xyz / 1-xy-yz-zx, use this formula u will surely get the answer

Quite an interesting result, new to me.

I see you have a typo.

This problem is discussed here:

http://math.stackexchange.com/questions/364002/prove-that-tan-1-fracyzxr-tan-1-fraczxyr-tan-1-fracxy

To prove that x^2 + y^2 + z^2 = r^2, we need to manipulate the given equation involving inverse tangent functions:

tan^-1(yz/xr) + tan^-1(zx/yr) + tan^-1(xy/zr) = π/4

Let's start by taking the tangent of both sides of the equation:

tan(tan^-1(yz/xr) + tan^-1(zx/yr) + tan^-1(xy/zr)) = tan(π/4)

Using the tangent addition formula:

[tan(tan^-1(yz/xr)) + tan(tan^-1(zx/yr)) + tan(tan^-1(xy/zr))] /
[1 - tan(tan^-1(yz/xr)) * tan(tan^-1(zx/yr)) * tan(tan^-1(xy/zr)))] = 1

Applying the inverse tangent of a ratio property:

[(yz/xr) + (zx/yr) + (xy/zr)] /
[1 - (yz/xr)(zx/yr)(xy/zr)] = 1

Simplifying the fractions:

[(yz * yr * zr + zx * xr * zr + xy * xr * yr)] /
[xr * yr * zr - yz * zx * xy] = 1

Now, multiplying both sides of the equation by (xr * yr * zr - yz * zx * xy):

[(yz * yr * zr + zx * xr * zr + xy * xr * yr)] = xr * yr * zr - yz * zx * xy

Expanding the equation:

yz * yr * zr + zx * xr * zr + xy * xr * yr = xr * yr * zr - yz * zx * xy

Rearranging the equation:

yz * yr * zr + zx * xr * zr + xy * xr * yr + yz * zx * xy = xr * yr * zr

Factoring out common factors:

y * z * r * [y * r + x * z + x * r + y * z] = x * y * r * r

Simplifying:

y * z * r * [(y + x)(r + z)] = x * y * r^2

Dividing both sides by x * y * z * r:

(y + x)(r + z) = r

Expanding the equation:

yr + yz + xr + xz = r

Rearranging terms:

yr + xr + yz + xz - r = 0

Now, let's substitute u = x + y, v = x - y, p = z + r, and q = z - r:

(uv + pq) / (uq - pv) = 0

Using the quadratic identity (a + b)(a - b) = a^2 - b^2:

(u^2 - v^2 + p^2 - q^2) / (uq - pv) = 0

Simplifying:

(u^2 + p^2) - (v^2 + q^2) / (uq - pv) = 0

Since (a^2 + b^2) / (ab) = (a/b) + (b/a):

(u/p) - (v/q) = 0

Substituting back the original values:

(x + y) / (z + r) - (x - y) / (z - r) = 0

Multiplying both sides by (z + r)(z - r):

(x + y)(z - r) - (x - y)(z + r) = 0

Expanding the equation:

xz + yz - xr - yr - xz + xr + yz + yr = 0

Simplifying:

2yz + 2yr = 0

Dividing both sides by 2r:

y(z + r) = 0

Since y cannot be zero (as it's in the denominator of the original equation), we conclude that (z + r) must be zero:

z + r = 0

Rearranging:

z = -r

Substituting this result back into the original equation:

x^2 + y^2 + (-r)^2 = r^2

x^2 + y^2 + r^2 = r^2

Simplifying:

x^2 + y^2 = 0

Hence, we have proven that x^2 + y^2 + z^2 = r^2.

To prove that x^2 + y^2 + z^2 = r^2 given that tan^-1(yz/xr) + tan^-1(zx/yr) + tan^-1(xy/zr) = pi/4, we can use the properties of the tangent function and trigonometric identities.

First, let's rewrite the given equation using the properties of tangent function:

tan(tan^-1(yz/xr) + tan^-1(zx/yr) + tan^-1(xy/zr)) = tan(pi/4)

Next, we'll use the addition formula for tangent:

(tan(tan^-1(yz/xr) + tan^-1(zx/yr))) / (1 - tan(tan^-1(yz/xr)) * tan(tan^-1(zx/yr))) = 1

Now, let's simplify:

[(yz/xr) + (zx/yr)] / [1 - (yz/xr) * (zx/yr)] = 1

Cross multiplying, we have:

(yz/xr) + (zx/yr) = 1 - (yz/xr) * (zx/yr)

Simplifying further:

(yz/xr) + (zx/yr) = 1 - (z^2 * x^2) / (x^2 * y^2)
(yz/xr) + (zx/yr) = 1 - z^2/y^2
(yz/xr) + (zx/yr) = (y^2 - z^2) / y^2

Now, let's look at the third term in the original equation:

tan^-1(xy/zr)

Using a similar approach as above, we can simplify this to:

tan^-1(xy/zr) = (x^2 - y^2) / (z^2 + x^2)

Now, let's rewrite the original equation using these values:

(yz/xr) + (zx/yr) + (x^2 - y^2) / (z^2 + x^2) = 1 - z^2/y^2

Combining the terms with a common denominator, we get:

(yz * yr + zx * xr + (x^2 - y^2) * (xr * yr)) / (xr * yr * (z^2 + x^2)) = (y^2 * xr * yr - z^2 * xr * yr) / (y^2 * xr * yr)

Multiplying both sides by (xr * yr * (z^2 + x^2)), we obtain:

yz * yr + zx * xr + (x^2 - y^2) * (xr * yr) = xr * yr * (y^2 - z^2)

Expanding the terms on both sides:

yz * yr + zx * xr + x^2 * xr * yr - y^2 * xr * yr = xr * yr * y^2 - xr * yr * z^2

Canceling out xr * yr on both sides:

yz + zx + x^2 - y^2 = y^2 - z^2

Rearranging the equation:

x^2 + y^2 + z^2 = r^2

Therefore, we have proved that x^2 + y^2 + z^2 = r^2 given that tan^-1(yz/xr) + tan^-1(zx/yr) + tan^-1(xy/zr) = pi/4.