MCQ
$\tan ({\cos ^{ - 1}}x)$ is equal to
- ✓$\frac{{\sqrt {1 - {x^2}} }}{x}$
- 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}} $
$\therefore \,\,\tan \,({\cos ^{ - 1}}x) = \tan \theta = \frac{{\sqrt {1 - {x^2}} }}{x}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to
|
X
|
-4
|
-3
|
-2
|
-1
|
0
|
|
P(X)
|
0.1
|
0.2
|
0.3
|
0.2
|
0.2
|