Question

If cos^–1x + cos^–1y – cos^–1z = 0, then show that x² + y² + z² – 2xyz = 1

Answer 1

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