MCQ
${\sin ^{ - 1}}\frac{1}{{\sqrt 5 }} + {\cot ^{ - 1}}3 $ is equal to
- A$\frac{\pi }{6}$
- ✓$\frac{\pi }{4}$
- C$\frac{\pi }{3}$
- D$\frac{\pi }{2}$
$ = {\cot ^{ - 1}}(2) + {\cot ^{ - 1}}(3) = {\cot ^{ - 1}}\left( {\frac{{2 \times 3 - 1}}{{3 + 2}}} \right) = {\cot ^{ - 1}}(1) = \frac{\pi }{4}$.
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$\text{e}^{-\text{x}}$
$\text{e}^{-\text{y}}$
$\frac{1}{\text{x}}$
$\text{x}$
$4\alpha=3\beta$
$3\alpha=4\beta$
$\alpha-\beta=\frac{7\pi}{12}$
$\text{none of these}$

$I$. $f$ is continuous on the closed interval $[a, b]$
$II.$ $f$ is bounded on the open interval $(a, b)$
$III.$ If $a$ $< a_1< b_1< b$, and $f (a_1)<0< f (b_1)$, then there is $a$ number $c$ such that $a_1 < c < b_1$ and $f (c)=0$