MCQ
${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right) = $
- A${\tan ^{ - 1}}x$
- ✓$\frac{1}{2}{\tan ^{ - 1}}x$
- C$2{\tan ^{ - 1}}x$
- Dએકપણ નહીં.
(Putting $x = \tan \theta )$
$ = {\tan ^{ - 1}}\left[ {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right] = {\tan ^{ - 1}}\left[ {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right]$
$ = {\tan ^{ - 1}}\left[ {\frac{{2\,{{\sin }^2}\frac{\theta }{2}}}{{2\,\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right]$
$ = {\tan ^{ - 1}} (\tan \frac{\theta }{2} ) = \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$.
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$A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]$
તો $A^{-1}$ મેળવો.
વિધાન $- II :$ ઉપર દર્શાવેલી ઘટના સ્વતંત્ર ઘટનાઓ છે.