# If cos–1x + cos–1y – cos–1z = 0, then show that x2 + y2 + z2 – 2xyz = 1 - Mathematics and Statistics

Sum

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

Concept: Trigonometric Equations and Their Solutions
Is there an error in this question or solution?