Maharashtra State BoardHSC Arts 12th Board Exam
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If cos–1x + cos–1y – cos–1z = 0, then show that x2 + y2 + z2 – 2xyz = 1 - Mathematics and Statistics

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Sum

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

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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
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