JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z + {\cos ^{ - 1}}t = 4\pi

, then the value of

{x^2} + {y^2} + {z^2} + {t^2}

View answer and explanation

Correct option: C

Option C (4) is correct because the given condition

{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z + {\cos ^{ - 1}}t = 4\pi

implies

x = y = z = t = -1

, and thus

{x^2} + {y^2} + {z^2} + {t^2} = 4

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)