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Question: If cos<sup>–1</sup> x + cos<sup>–1</sup> y + cos<sup>–1</sup> z = p, then-...

If cos–1 x + cos–1 y + cos–1 z = p, then-

A

x2 + y2 + z2 + xyz = 0

B

x2 + y2 + z2 + 2xyz = 0

C

x2 + y2 + z2 + xyz = 1

D

x2 + y2 + z2 + 2xyz = 1

Answer

x2 + y2 + z2 + 2xyz = 1

Explanation

Solution

Ž cos–1(x) + cos–1(y) + cos–1(z) = cos–1 (–1)

Ž cos–1(x) + cos–1(y) = cos–1(–1) – cos–1 (z)

Ž cos–1(xy – ) = cos–1{(–1)(z)}

Ž xy –= –z

squaring both sides we get

x2 + y2 + z2 + 2xyz = 1

Trick : Put x = y = z = 12\frac { 1 } { 2 }

so cos–112\frac { 1 } { 2 }+cos–112\frac { 1 } { 2 } + cos–112\frac { 1 } { 2 } = p

Obviously (4) holds for these values of x, y, z.