MCQ
$\tan ({\cos ^{ - 1}}x) = . . ..$
- ✓$\frac{{\sqrt {1 - {x^2}} }}{x}$
- B$\frac{x}{{1 + {x^2}}}$
- C$\frac{{\sqrt {1 + {x^2}} }}{x}$
- D$\sqrt {1 - {x^2}} $
$ \Rightarrow \,\,\tan \theta = \sqrt {{{\sec }^2}\theta - 1} $
$ = \sqrt {\frac{1}{{{x^2}}} - 1} = \sqrt {\frac{{1 - {x^2}}}{x}} $
$\therefore \,\,\tan \,({\cos ^{ - 1}}x) = \tan \theta = \frac{{\sqrt {1 - {x^2}} }}{x}$.
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