If ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,\;{\rm{then}}\;{x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right)$ Where $k$ is equal to

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Analyzing +4 | -1

{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,\;\mathrm{then}\;{x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right)

Where

k

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because after applying trigonometric identities and simplifying the given equation, the value of k is found to be 2.

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