MCQ
Write $\cot ^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right), x>1$ in the simplest form.
- ✓$\sec ^{-1} x$
- B$cosec ^{-1} x$
- C$tan ^{-1} x$
- D$cot ^{-1} x$
Therefore, $\cot ^{-1} \frac{1}{\sqrt{x^{2}-1}}=\cot ^{-1}(\cot \theta)$$=\theta=\sec ^{-1} x$
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$($ where $det(B)$ denotes determinant of Matrix $B) -$
$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$
Where $a, b \in Z$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to ..........