MCQ
${d \over {dx}}[{\tan ^{ - 1}}(\cot x) + {\cot ^{ - 1}}(\tan x)] = $
- A$0$
- B$1$
- C$-1$
- ✓$-2$
=$\frac{{1( - {\rm{cose}}{{\rm{c}}^2}x)}}{{1 + {{\cot }^2}x}} - \frac{{1({{\sec }^2}x)}}{{1 + {{\tan }^2}x}} = - 1 - 1 = - 2$ .
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$ ની કિમંત મેળવો.