[ ^ - 1 1 + x^2 ] = - Vidyadip

MCQ

$\tan \left[ {{{\sec }^{ - 1}}\sqrt {1 + {x^2}} } \right] = $

A
$\frac{1}{x}$
✓
$x$
C
$\frac{1}{{\sqrt {1 + {x^2}} }}$
D
$\frac{x}{{\sqrt {1 + {x^2}} }}$

✓

Answer

Correct option: B.

$x$

b
(b) $\tan \,\left( {{{\sec }^{ - 1}}\sqrt {1 + {x^2}} } \right) = \tan \,\left( {{{\sec }^{ - 1}}\sqrt {1 + {{\tan }^2}\theta } } \right)$

(Putting $x = \tan \theta )$

$ = \tan \,({\sec ^{ - 1}}\,\sec \theta ) = \tan \theta = x$.

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