If ^{-1} x + ^{-1} y + ^{-1} z = π then:

Correct Answer is : x2+y2+z2=1−2xyzx² + y² + z² = 1 - 2xyzx2+y2+z2=1−2xyz

Correct Answer is : x2+y2+z2=1−2xyzx² + y² + z² = 1 - 2xyzx2+y2+z2=1−2xyz

Test Index

UPSC NDA Math Model Paper 7

Question : 64 of 120

Marks: +1, -0

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

Solution:

Given:

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

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

\Rightarrow \cos^{-1}[xy - \sqrt{(1-x^{2})(1-y^{2})}]

= \pi - \cos^{-1} z

\Rightarrow [xy - \sqrt{(1-x^{2})(1-y^{2})}]

= \cos(\pi - \cos^{-1} z)

\Rightarrow [xy - \sqrt{(1-x^{2})(1-y^{2})}]

= -\cos(\cos^{-1} z)

(∵ cos (π - θ) = - cos θ)

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

(∵ cos(cos

^{-1}

x) = x)

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

Squaring both sides, we get

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

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

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

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

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

Go to Question:

Prev Question

Next Question