MCQ
${d \over {dx}}\sqrt {{{1 - \sin 2x} \over {1 + \sin 2x}}} = $
- A${\sec ^2}x$
- ✓$ - {\sec ^2}\left( {{\pi \over 4} - x} \right)$
- C${\sec ^2}\left( {{\pi \over 4} + x} \right)$
- D${\sec ^2}\left( {{\pi \over 4} - x} \right)$
$ = \frac{{1 - \tan x}}{{1 + \tan x}} = \tan \left( {\frac{\pi }{4} - x} \right) $
$\Rightarrow \frac{{dy}}{{dx}} = - {\sec ^2}\left( {\frac{\pi }{4} - x} \right)$.
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વિધાન $1:$ $S=\{x:f(x)=f^{-1}(x)\}=\left\{ {0, - 1} \right\}$
વિધાન $2$ : $ f $ એ એક-એક અને વ્યાપત છે.