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.