JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

0 < x < 1

, then

\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cot }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}}^2} - 1} \right]^{1/2}}

is equal to

View answer and explanation

Correct option: C

x\sqrt {1 + {x^2}}

Option C (

x\sqrt {1 + {x^2}}

) is correct because simplifying the expression using trigonometric identities leads to this result.

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x

{x^2} + {y^2} + {z^2} = {r^2},

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