If p > q > 0, pr < - 1 < qr, then prove that tan^-1((p - q)/(1 + pq)) + tan^-1((q - r)/(1 + qr)) + tan^-1((r - p)/(1 + rp)) = π

Question 1

If p > q > 0, pr < - 1 < qr, then prove that tan-1((p - q)/(1 + pq)) + tan-1((q - r)/(1 + qr)) + tan-1((r - p)/(1 + rp)) = π

Question 2

We have

tan-1((p - q)/(1 + pq)) + tan-1((q - r)/(1 + qr)) + tan-1((r - p)/(1 + rp)) = π

= (tan-1p - tan-1q) + (tan-1q - tan-1r) + p + (tan-1r - tan-1p)

= π

kratos · Answer 1 · 2021-01-06T20:20:57+0000

We have

tan-1((p - q)/(1 + pq)) + tan-1((q - r)/(1 + qr)) + tan-1((r - p)/(1 + rp)) = π

= (tan-1p - tan-1q) + (tan-1q - tan-1r) + p + (tan-1r - tan-1p)

= π

Please log in or register to add a comment.