If cos^{–1}x + cos^{–1}y – cos^{–1}z = 0, then show that x^{2} + y^{2} + z^{2} – 2xyz = 1

#### Solution

cos^{–1}x + cos^{–1}y – cos^{–1}z = 0 .......[Given]

∴ cos^{–1}x + cos^{–1}y – cos^{–1}z

Let,

cos^{–1}x = M and cos^{–1}y = 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")` .......[∵ sin^{2}θ + cos^{2}θ = 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^{–1}z = `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 – x^{2})(1 – y^{2}) = (xy – z)^{2}

∴ (1 – x^{2})(1 – y^{2}) = x^{2}y^{2} + z^{2} – 2xyz

∴ 1 – x^{2} – y^{2} + x^{2}y^{2} = x^{2}y^{2} + z^{2} – 2xyz

∴ x^{2} + y^{2} + z^{2} – 2xyz = 1