MCQ
$1 + {\cot ^2}({\sin ^{ - 1}}x) = $
- A$\frac{1}{{2x}}$
- B${x^2}$
- ✓$\frac{1}{{{x^2}}}$
- D$\frac{2}{x}$
Now $1 + {\cot ^2}\theta = \cos e{c^2}\theta = \frac{1}{{{x^2}}}$
Hence $1 + {\cot ^2}\,({\sin ^{ - 1}}x) = \frac{1}{{{x^2}}}$.
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| $Face:$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| $P(F)$ | $0.1$ | $0.24$ | $0.19$ | $0.18$ | $0.15$ | $0.14$ |
If an even face has turned up, then the probability that it is face $2$ or face $4$, is
A coin is tossed 4 times. The probability that at least one head turns up is:
$\frac{1}{16}$
$\frac{2}{16}$
$\frac{14}{16}$
$\frac{15}{16}$