If cos–1x + cos–1y – cos–1z = 0, then show that x2 + y2 + z2 – 2xyz = 1
Solution
cos–1x + cos–1y – cos–1z = 0 .......[Given]
∴ cos–1x + cos–1y – cos–1z
Let,
cos–1x = M and cos–1y = N
∴ M + N = cos–1(z) .......(i)
Also,
x = cos M and y = cos N .......(ii)
∴ sin M = `sqrt(1 - cos^2"M")` and sin N = `sqrt(1 - cos^2"M")` .......[∵ sin2θ + cos2θ = 1]
∴ sin M = `sqrt(1 - x^2)` and sin N = `sqrt(1 - y^2)` .......(iii)
Consider
cos(M + N) = cos M cos N – sin M sin N
∴ cos(M + N) = `xy - sqrt(1 - x^2) sqrt(1 - y^2)` .......[From (ii) and (iii)]
∴ M + N = `cos^-1(xy - sqrt(1 - x^2) sqrt(1 - y^2))`
∴ cos–1z = `cos^-1(xy - sqrt(1 - x^2) sqrt(1 - y^2))` .......[From (i)]
∴ z = `xy - sqrt(1 - x^2) sqrt(1 - y^2)`
∴ `sqrt(1 - x^2) sqrt(1 - y^2)` = xy – z
Squaring both sides, we get
(1 – x2)(1 – y2) = (xy – z)2
∴ (1 – x2)(1 – y2) = x2y2 + z2 – 2xyz
∴ 1 – x2 – y2 + x2y2 = x2y2 + z2 – 2xyz
∴ x2 + y2 + z2 – 2xyz = 1
