If cos^-1 p cos^-1 q cos^-1r=pi then prove that p^2 q^2 r^2 2pqr=1?

Question

Answer

Proof:

We are given that,

cos^-1 p cos^-1 q cos^-1 r = π

Let's assume that,

cos^-1 p = A, cos^-1 q = B, cos^-1 r = C

Then we have,

p = cos A, q = cos B, r = cos C

Now, using the formula cos(A + B + C) = cos A cos B cos C - sin A sin B sin C, we get

cos(A + B + C) = pqr - √[(1 - p^2)(1 - q^2)(1 - r^2)]

Since cos(A + B + C) = -1, we have

-1 = pqr - √[(1 - p^2)(1 - q^2)(1 - r^2)]

Squaring both sides, we get

1 = p^2 q^2 r^2 - 2pqr√[(1 - p^2)(1 - q^2)(1 - r^2)] + (1 - p^2)(1 - q^2)(1 - r^2)

Now, we need to prove that p^2 q^2 r^2 2pqr = 1.

Multiplying both sides of the earlier equation by 2pqr, we get

2pqr = 2p^2 q^2 r^2 - 2pqr√[(1 - p^2)(1 - q^2)(1 - r^2)] + 2pqr(1 - p^2)(1 - q^2)(1 - r^2)/2pqr

Since cos^-1 p, cos^-1 q, and cos^-1 r are all less than or equal to π/2, we have p, q, and r all greater than or equal to 0. Therefore,

√[(1 - p^2)(1 - q^2)(1 - r^2)] ≤ √[1(1)(1)] = 1

So, the last term in the above equation is less than or equal to 1. Therefore,

2pqr ≥ 2p^2 q^2 r^2 - 2pqr + 2pqr(1 - p^2)(1 - q^2)(1 - r^2)/2pqr

Cancelling out the 2pqr terms, we get

1 ≥ p^2 q^2 r^2 + (1 - p^2)(1 - q^2)(1 - r^2)

Expanding the last term, we get