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MHTCET 10 May 2024 Shift 1 Paper

Question : 144 of 150

Marks: +1, -0

\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi

x^{2}+y^{2}+z^{2}+k x y z=1

, then k is

Solution:

Concept:

Use the standard formula for the sum of two inverse cosines and then compare the resulting equation with the given one.

Explanation:

Let

A = \cos^{-1}x

B = \cos^{-1}y

, and

C = \cos^{-1}z

Given that

A + B + C = \pi

, so

A + B = \pi - C = \cos^{-1}(-z)

Taking cosine on both sides:

\cos(A+B) = -z

Using the identity

\cos(A+B) = xy - \sqrt{1-x^{2}}\sqrt{1-y^{2}}

, we get:

xy - \sqrt{(1-x^{2})(1-y^{2})} = -z

Rearranging:

xy + z = \sqrt{(1-x^{2})(1-y^{2})}

Squaring both sides:

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

Simplifying:

x^{2}y^{2} + 2xyz + z^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}

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

Comparing with

x^{2} + y^{2} + z^{2} + kxyz = 1

, we obtain

k = 2

Answer:

k = 2

, which is option D.

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